Epsilon-Delta Definition of Limits: Solving Problems

in this video we solve several limit problems using the epsilon delta definition uh you may watch this video after watching the oge video that discuss one problem and also the theory video that explain the theory behind the epsilon delta definition you can find the link in the description so in this video are we gonna solve uh several problems in different ranges for example we have linear problems and quadratic and quadratic with the two and three terms and cubic and square root functions uh rational functions and then also square root in rational form so these are all type of problems we have nine problems here so let's see how we solve each one so i'm going to use a different logical arguments like uh sometimes double implicit implications and sometimes single implicit implications and also sometimes like a simple calculations so let's see how we can solve each one so let's start with the uh first let's start with the definition so the precise definition so what's a precise definition so we say that the limit when x goes to a f x equal l are if and only if but because a double implication if and only if for all epsilon that means for any given epsilon there is a delta which is a function of epsilon such that for all x in the domain of the function because this is very important especially when there is a discontinuity so for squared functions or rational function you have to be very careful with this one ah then t 0 less than x minus a less than delta implies uh the effects the difference between the function and the limit is less than epsilon so if you can prove that for all epsilon there's a delta and for all x in the domain uh the first one implies the second one that means if x is within delta this distance then you want to make sure that the function is within the l uh the functions function is within the epsilon distance to the l so so that's the what we call the precise diffusion or the limit definition good so uh so let's look at this problem so we're going to go to the first problem so what it says that so we have um so prove that prove that the limit when x goes to uh let's say eight just pick a number a one half x minus three since you're more terms you're able to parent this around equal one that's what we're trying to do we have this one coming from and you can see that one half x minus 3 is a linear function this is a polynomial so that means you can just plug in the value to find because these are continuous functions so you can plug in the value you plug in eight and you can see that you get one so that's where the one coming from but how we so what we're trying to do right now is we're gonna prove that that's exactly the result so uh so to do uh so to prove a problem like this what we're going to do so we can use the epsilon definition so we pick a general epsilon so any epsilon and then we try to find a delta that implies uh this condition so uh so let's see how the argument gonna work so the proof so what we can do we're gonna pick a uh epsilon let's say let epsilon positive so you pick a positive epsilon and then what we need to do so need to show that so let's see how to write it like a logical complete logical proof okay so i need to show that need to show that there is a delta there is a there is a delta which is a function of epsilon let me depend on epsilon such that such that 0 x minus 8 less than delta implies the function the function is one half x minus three minus the limit in this case it is one actually minus one that's the element we are talking about 1 is less than epsilon so the goal is to find the delta so and then you can see that this is the function is the function and that is l so this is uh l so what we normally do is uh because we don't know what the delta is so we we start with what we need to prove and then think about the condition and go forward so what we do so we can see now are we gonna pick what we want to prove so what you want to prove is one half x minus three minus one less than epsilon that's what we want to prove so and then we we know that this if and only if this double implication we're going to simplify inside so if you simplify inside what you get one half x minus 4 which is less than epsilon so they are they're equivalent now what we're going to do we're going to multiply both sides by one half so that means this is if and only if you multiply inside like both sides you can just uh bring inside by uh by 2. why we do that your goal is to create the x minus eight term now when you multiply you can see this is again if and only if x minus or you're gonna get x minus eight less than a two epsilon so you can see that this is exactly what we want to prove so what we're going to do so this can give an option so you can see that if you pick this part as delta then this implies that i implies uh the condition so that means this simply gonna give you uh the delta so one good example uh option would be delta equal to epsilon you really don't want to do that because you can pick any delta which is smaller than this so one what do you say optimal option would be here it's not the optimal option so would be here two epsilon but you can pick any delta which is smaller than two epsilon for example you can pick delta equal epsilon that still works because because delta is the uh uh the gap or the radius uh from eight so that means even if you pick a smaller one that still works but that's still within that so that means you can pick any smaller um delta for this one but we're gonna go with this one because that's what we directly get from this one but again you can pick anything so now we can say so thus so it's very simple now argument does because it's only final leaf that means if you start with one you get the other one so that's a choosing just choosing delta equal to epsilon or you can pick any for example you can pick the like epsilon that we can do later in another example so we have we have because of this all double implications you can say 0 less than x minus 8 are less than delta implies now we can write just all like one direction for example you can write this one go this way and then this one go that way and all go like a backward only only like a left so then you can you get the answer so this implies that uh the first one so you can say one half x minus three minus l in this case the 1 are less than epsilon that's exactly what you want to prove so there's a function and this is the limit so this completes the proof this is the epsilon delta argument so this complete because if you pick any epsilon and you prove that the iso delta that works okay so this completes the proof this completes the proof good so you can see that's very simple it's not that difficult but there are some logical arguments that you want to make so let's look at the next problem so as the next one we have proved that limit x goes to 2 x cube equal eight again uh x cube bc polynomial so you can plug in the value and get the uh limit so which is eight but let's see how we can prove this so in this time i'm gonna use a slightly different argument and let's see how it works so the proof proof let's say let epsilon be positive so you're picking a positive epsilon ah then note that note that uh so we can start what we want to prove we're going to start with x 3 minus 8 equal and we know that we can factor this how to factor this so you can see the 8 means 2 to the 3. so we you can use the difference of cube formula we know that a3 minus b3 equal a minus b the first factor the second factor is a squared plus a b plus b squared so that is how you're going to factor cubes since uh 8 is 2 q we can use that so because with that if you look at those two factors we can write this like this the absolute value the first factor minus times the second factor you can add it like that because it's a product plus 2x plus so this is true they are equal now what we're going to do so the goal is to find the log because x minus 2 that so we need x minus 2 that's we already have so the goal is to find the largest value of this factor so find the largest value of the rest of the function find the rest of the expression find the largest value that's how we need largest ready find the largest value but the problem is we don't know what's the largest value because this is just a polynomial it can have any any value because it goes to infinity so there's no largest value here so that means we need to restrict uh the domain so what we're gonna do we're gonna restrict x minus two so we're gonna say um so we can start with like uh so um so let x minus 2 is within 1. so we're going to pick values you can pick any value you like so let's say 1. so that means what that means if you look at this interval what this means it means that the all the values which are one unit away from two so that means uh you add once you get three and you subtract you get one so those are all the values that you're gonna get since we are looking for the largest value i'm going to put the largest value here so pick the largest value if you're the largest cell in this case 3 is it's not in there but we are looking for a bound so that means you can say 3 here what we can do we're going to see what will happen this this that factor so if you put x equal 3 x squared means 9 plus 2 times 3 means 6 plus 4 that means this simplifies to 19 so this you can see that the largest possible value under this condition that you can get is 19 so that means uh but if you restrict uh this one to delta so if you restrict this sort of delta let's say you this is smaller than delta that means the largest you can get here is 19 delta because the other one gives you 90. so what your goal is to find the largest value of the other factor so the condition is you want to make sure that this is smaller than epsilon so what we're going to do we're going to pick a really smaller delta such that 19 times delta you don't need 90. you can say 20 here for the safe side but 19 is the largest value you can get so you what you can do you're going to pick a really small delta so that the 19 delta is smaller than epsilon so you can always do that you can pick a really really small delta that means you can make sure that 19 delta is smaller than epsilon actually you can even have equal signs so we already have a strict inequality so that's that's exactly how you're going to find delta so after you get that you want to make sure that you're going to pick a really smaller delta such that 19 delta is smaller than or equal to epsilon so that means now we're going to solve this so if you solve this you can see this is 19 uh delta smaller than epsilon this is the delta is less than or equal to is epsilon over 19. you can pick any like for example you can pick epsilon or 20. so anything works so it's 19 or more now we have all the conditions we need so that means now what happened so so uh we have now and so the condition is and so this says that x minus ah 2 less than epsilon over 19. now what we're going to do you want to make sure that both conditions are satisfied so we're going to pick pick a delta which is the smallest of the two what are we going to do so so what we're gonna now choose so this interesting part now choose delta which is the minimum of those two minimum of 1 or and epsilon 90 because if you pick the minimum then it's guaranteed that it satisfy both of them so now let's see how the rest of the argument works so then so you can write so you can write this then you can say if epsilon 0 less than x minus in this case 8 actually uh it is 2 so 2 less than delta this implies that because since delta is the minimum this implies that x minus 2 less than any of them so that means for example epsilon 90 will be the minimum since we pick the minimum it is smaller than any of them now but we're going to multiply by 19 both sides so it's x minus 2 times 19 less than epsilon but we know that since 90 is the largest value we can bring that back a factor back here so we can write we can bring the other factor here so it is x squared plus 2x plus 4 that's the other factor which is less than epsilon but now we can combine them and we know that this is this side is smaller than 19 so this implication works so this implication is true because 90 is the largest value so if you replace something by something smaller then still it should be valid now we're going to combine them you're going to multiply through once you multiply through so you could write this one i would write the extra step here you really don't need that and you can bring into one product like that now you can see that because of the uh difference of cubes argument this simply equals to x 3 minus 8 less than epsilon and that's exactly what we will produce that's exactly how the argument works so it's directly implies uh that way so that means so thus what we prove so we prove that for all epsilon there is a delta which is a which is depend on epsilon such that 0 x minus 2 less than delta implies x 3 minus 8 less than epsilon that's exactly the definition so therefore by the definition let's have you write the argument therefore by the definition therefore by the definition the limit x goes to 2 x cubed equal 8. so this completes the proof so you can see this is a really nice neat argument so let's look at the next example so as the next example we have proved that the limit when t goes to three t plus one square root equal to um so we're gonna we prove and we we notice something here we see that this is called function we wanna make sure this part is greater than or equal to 0 so this implies t greater than equal negative 1 so that means you want to make sure that you are not going to plug in anything uh which is smaller than negative 1 let's see how we can take care of that later so again uh so we're gonna use a similar argument so we're gonna say let ah epsilon greater than zero uh then note that know that so what we're going to do we're going to start with what we need to prove note that t plus 1 square root minus 2 that's the one we're going to consider subtract the limit from the function so what we're going to we're going to simplify this we are not going to write anything with epsilon here we just simplify this expression so this is equals what we're going to do when you get square root what we normally do is we're going to multiply by a special one where the special one is the conjugate top time bottom the numerator and the denominator so that's the argument we're going to use here so that means we have t plus 1 square root minus 2 times you're going to multiply the top and bottom by the conjugate conjugate means you change the sign here in between not inside change the sign for the subtraction so uh so go plus sign and then also we do that with the minus sign so so this is simply 1 so you can always multiply by 1 so we do that and it's under ah absolute sign now you can see that when you uh multiply the top we can use this result we know that if you have a minus b with a plus b you're gonna get a squared minus b squared that's exactly the reason why we multiply by the conjugate so so one is the conjugate of the other so and because of that you're going to square the first one so when you square the first one you'll get t plus one minus minor second one you're going to get 4 so that's what you get for the top for the bottom and you can see that this is always greater than or equal to 0 x is greater than or equal to 1 um this is square root so it's always we assume that's greater than equal to 0 so because of that this whole expression is positive so that means you can so we still have the absolute value for the top but for the bottom you can take it out because that's a positive term so plus 2 and then so that's what you get now uh you can see this again simplifies to the top you know simplifies to t minus 3 so that's exactly the term that we want to bound and the other thing i'm going to write as a separate term uh because then the argument would be much easier so we have this so now what we're going to do is we're going to make you want to find the largest value of this expression you'll find the largest value of this expression but this one is very clear because we know that the t plus one square root is greater than or equal to zero so that means the largest value you get when this is zero just that's when you get the largest value uh so that means you can see that this whole expression is less than or equal to one-half so you can write it like that so that means you can write this on s so this is less than or equal to you can see this is less than or equal to uh but if you if you bound uh this one by delta because this is smaller you can see that this is smaller than one-half delta one half coming from uh the other expression we know that this expression the whole expression is less than equal one half one half you get uh the t when t equals negative one that's when you get uh one half otherwise this is smaller because you are dividing the numerator by something so that means this is smaller so always positive that is we add a positive term to the new denominator that means the fraction gets smaller so um and then and then uh so what we're going to do that we're going to pick a small delta so that this one-half delta is smaller than epsilon that's what we're trying to do so we're going to pick a delta smaller delta which is uh so that one-half delta is smaller than epsilon so that means if you solve for delta you can see that delta is smaller so what you get from here you see that delta is smaller than ah less than equal to 2 epsilon so that means you can pick either two epsilon or any other epsilon uh any other uh function delta which is smaller than uh two epsilon for example delta equal epsilon but uh but there's something you have to be careful because you know that you want to make sure that you do not pick any values of t which are less than uh negative 1. to guarantee that what we're going to do we're going to put extra condition on delta so you say that so now what we're going to do now choose this interesting part now choose now choose delta equal the minimum of just put a number here like one or one half something like that so that it doesn't go until ah negative one so you can see that if you uh put a delta like that uh so you can see what we're trying to do is we try to go this is three so you start to stay around this interval four and two you can see that the negative one is is far away so that means it's never reached negative one that means there is no any discontinuity here like there are no any undefined point this is exactly what how to do that we just put a um some number so that doesn't go that far that's why i put one here if you don't put that there the argument wrong and then we have and then we get the minimum of 1 and 2 epsilon so once you do that then you can write a 0 less than a t minus 3 less than delta implies that um t minus 3 less than any of them so for example 2 epsilon because the minimum will repeat the delta as the meaning of c is the minimum it is smaller than any of this more than one also smaller than the other one two epsilon now what we can do we're gonna divide both sides by two so we can write this one as t minus three times one half dividing both sides by 2. now what we can do we know that one-half is the largest value of the expression so that means we can put that back in um so we're going to put that back in so we say t minus 3 times we're going to put the expression here so what is the expression it is t plus 1 square root plus 2 less than epsilon why because we know that that expression is bigger than a one-half one have the largest values if you put something smaller it is small uh now what we can do are we gonna write backward and we know that this is if and only if uh you put the multiplied by the conjugate so this is fine only so they are equal so what we can do t minus 3 we can write we can break it into t plus 1 square root minus 2 times t plus 1 square root plus 2. so the benefit is because we know that this is exactly steps that is that's why we can write this like otherwise it's going to be like very surprising where is coming from but we know that this is exactly how we simplify this uh so now what we're going to do we're going to combine uh you can see you can cancel this with that so once you cancel that so this is again different only because they're all equal so if an only if you can write this one as t plus 1 square root minus 2 less than epsilon so now we have the proofs that's exactly what we want to prove so therefore you can say therefore by the definition by the definition epsilon delta diffuse by the definition you prove that the limit t goes to r r3 t plus 1 square root equal to so that completes the proof so let's look at the next problem so we have uh prove that limit x goes to 2 3 x square plus 4 equals 16 so we're going to use a similar argument so the proof uh so let epsilon be positive and then say consider consider so we're going to consider what we need to prove sorry 3x squared plus 4 a minus 16. but we know that this is equals to if you simplify you can see that you're gonna get 3x squared minus 12 that's what you get and then what we can do is we can factor we can pull 3 out so if you pull 3 out as a factor you're going to get x squared minus 4. now we can factor x square minus 4 using a different squares so that means what we can do is we can write the x minus 2 here and we separate the rest of the rest from this one so we do the multiplication three times x plus two and then you can also combine and you can write it like this so we're going to say x minus 2 you're going to combine it and we're going to put 3 x plus 6 we're going to combine like that and now the goal is to abound the second term so to do that we do as we because we know that this there's no uh otherwise there's no largest value of that expression so we define the largest value of this expression but there's no largest value it goes infinity so what we're going to begin about the delta so what we do as before we're going to say are we going to pick a value x minus 2 less than 1 because that's going to rescue the x value so this says that x is within 1 that means largest value is 3 and the smallest value is 1. since we are looking for the largest value uh we're gonna pick so that's the largest value three so we're going to use that value uh so that means if you use three and you can see that it's a three times x uh so it's 3 times 3 plus 9 that means this is less than 19 because this is an open interval so that means what we what we will make sure that is so we will make sure that this is less than epsilon so what we're going to do we're going to find a delta we're going to find a delta which is smaller than in this case actually you get 15 so it's not 90 50. so um so we're gonna we're gonna find a delta so that the 15 times delta is smaller than epsilon uh because we make sure that this is less than delta so that's where the 15 coming from so 15 times delta we make sure that we finally a smaller delta so that 15 delta is smaller than epsilon that you can always do so pick a delta like that um so now we can write the rest of the argument so now we can say now choose so now i chose a delta which is the minimum of uh it's a minimum of one and so you can see when you simplify for delta you're gonna get uh delta is less than epsilon over 15. so you can pick this or any smaller values so uh so we got epsilon of 50. so you can say 720 that works so then what happened then because of this equality and you can see that then 0 less than x minus in this case 2 less than delta implies that x minus 2 times x minus 2 absolute x 1 is 2 less than we can put epsilon over 15 so this implies x minus 2 times 15 less than epsilon now what we're going to do we're gonna replace 15 by the expression because if the expression says that is smaller the implication works so what we do instead of 15 now we can write 3x plus 6 less than epsilon that implies that that directly implies that what we want since the um the left hand side is equals to this one so you can write this one as 3 x squared plus 4 minus 16 less than epsilon because they are all equal so that completes the proof this completes the proof so that's how you make the a logical argument for for a problem like this so let's look at the next one so as the next one we have limit x goes to 1 4 plus x minus 3x squared equal to so we do exactly the same way we did before so the proof so we're going to say let epsilon are positive then what we can do is we can write several implications so we can start with what we want to prove so it is for plus x minus 3 x squared a minus 2 which is less than epsilon that's having proof this implies like it's like double implication this if and only if uh but we're gonna we're gonna simplify the left hand side so if you simplify you're gonna get 2 plus x minus 3 x squared are less than epsilon this is if and only if you know that inside the absolutely you can switch the order so we can write this one as 3x squared so you can simply multiply by negative sign so 3x squared and then minus x minus 2 that's exactly the same you can only switch inside the absolute sign now we can factor so you can see that this says factors is again if an only if they are all equal x minus 1 times the other factor is 3x plus 2 less than epsilon now this is again if and only if you can write this one as if you pull the x minus 1 out or you get the other factor which is 3x plus 2 3x uh plus 2. so now what we want to do is to find the largest value of this one so this is very similar to the previous one what we're going to be going to bound x minus 1 first so we can say that so we're gonna bound that so we use the same trick every time so this is around one so we're gonna go uh we're gonna pick x minus one uh less than one again so that's what we're trying to do so that means it's going to go up to 2 and 0. so 0 and 2 since we are looking for the largest value we pick this side because we pick 2 here so that's the largest value so we're going to pick that one so that means then we can see that uh if we the largest values x equal 2 so you'll get 3x plus 2 equals 8. so that means now we have only single implication it's only go left side so this you can see that then we can write if you pick the largest value we have implications to the left side so it's gonna see repeat the larger that's the largest value of three x plus two uh so it's eight uh less than so what we can do we wanna make sure that now we're going to pick a delta so that 8 times delta is smaller than epsilon so that means we're going to pick a we can now we're going to pick a smaller delta so you look at this smaller delta so that 8 times so that 8 times delta so it doesn't both looks like 8 so it is 8 times delta is less than or equal to epsilon that means if you simplify for delta you're going to see that delta is are less than equal epsilon over eight so that means then what we can do we can pick a value that satisfy both conditions so that means now we say choose choose delta equal the minimum of the two minimum of one and epsilon number eight so then we have this uh simplification so we can see that because of this double implication we can simply write 0 less than x minus 1 less than delta implies because the double implication we can say x minus 1 less than epsilon over eight so complicating because that's the smallest now um this implies that uh four plus x minus three x squared minus 2 less than epsilon so that completes the proof so as the next one we have limit i u goes to 1 3 over u plus 2 equal 1 and you can say this is a rational function so you want to make sure that you do not include the value uh negative two with that value gonna give you something undefined so uh let's see how it works so the proof uh so let epsilon be positive then you can write this statement you can say a 3 over u plus 2 minus 1 that's the difference between the functional limit less than epsilon if and only we're going to simplify inside so if you simplify inside that we're going to do cross multiplication you can write this on as 3 minus u plus two over r u plus two uh less than epsilon so this is final leaf and again this is phenol leaf you can see at the top you can simplify as 1 minus u absolute value and then again u plus 2 absolutely since we don't know what is positive so we're going to leave it like that and then you can also switch the top because it's on the absolute sign so you can switch the top and also isolate this term u plus 2 absolute value now your goal is to find the largest value of this expression that's not that difficult so what we can do as before we can restrict the domain for u so that means so let's do that uh so u is around 1 and then again so we're gonna say u minus 1 are less than 1 not x so u minus 1 are less than 1 so when you do that then you're going to see that this is 2 and 0. so now so let's see that so we're going to pick the values in this interval so we're going to pick the values only in this interval but we are looking for the largest possible value since this is in the denominator now we're going to pick the value on the other corner because this is the value we're going to pick so now we pick the smallest value smallest 20. it will be the smallest value you get the largest fraction so once you plug in the smallest value that means 0 and you can see that you can write this sentence so we can see so this is uh left implied that u minus 1 absolute value now you're going to pick the largest value with strip with the largest value then it's going to give you a left implication so ah less than now again we restrict this side u minus 1 by delta so that means uh you can see that we want to make sure that this is within the less than epsilon so what we can do we're going to pick a value a delta so that the half times delta we have is the largest possible value you can get from the other expression so ah the half delta so you pick a smaller delta so that half delta is smaller than epsilon so it's very clear so that means uh and then we're gonna we're gonna use this to simplify for delta so this is the delta is smaller than two epsilon so we're gonna choose a delta which is smaller than one and two epsilon so now uh so it's as before so you can see that uh choose so i'm going to show you a different argument this time so let's say choose because you know that you can pick either a delta equal arms it's a less than or equal so it's a delta equal uh 2 epsilon or any any smaller one so let's say for example you're going to pick delta equal epsilon this time and see how the argument works so delta equal minimum so you can pick either minimum of 1 and 2 epsilon or even any other estimator so let's let's say you pick epsilon in this case so let's see how the argument works now so then you can write this then 0 less than u minus 1 less than delta implies that u minus 1 is less than epsilon because that's how we pick the delta which has to be smaller than uh any of them but you know that this is smaller than 2 epsilon so that's how you're going to jump through this step now what we can do we're going to skip the epsilon we're going to go with the 2 epsilon so that means this implies now you can say that u minus 1 uh i'm going to divide by half so i'm going to say okay let's try this step so this is true now now what we can do we can divide by 2 both sides and then since epsilon one half is the largest value this implies u minus one so we're going to put u uh plus two absolute value back in there now you're going to combine so you can see that because of the argument we write because the double implication we wrote this is exactly similar to 3 over u plus 2 minus 1 that's exactly similar less than epsilon so that means that's exactly what we want to prove so this completes the proof this completes the proof so you can see that it doesn't take much time and if you understand the how the argument works and so you can see that we can even pick a small epsilon but still the argument works good so let's look at the next one so as the next one we have limit t goes to negative 1 6 minus 3 d square root equal 3 um so we're going to do a similar argument here but we're going to use the simplification step now no no uh implications so let's say proof so we pick a epsilon so let epsilon be positive so p capacity epsilon and note that so i'll write different type of proofs so that you can learn different arguments so let's say we can start with what we want to prove 6 minus 3 t square root minus 3 that's the limit and the function this is equals to again we're gonna multiply the uh top and bottom by the so we have six minus three t square root minus 3 that's the first term times we're going to multiply the conjugate of this so we only change the sign between not inside so that means we have 6 minus three t square root plus three and we divide by the same thing six minus three t square root plus three uh no epsilon here now what we can do we're going to combine the top so it's because of the differential squares formula you can see the top is now 6 minus 3t minus 9 and we would absolutely sign there and then the bottom bottom you can see 6 minus 3t square root plus 3. we don't need the absolute sign the reason is this term is greater than equals 0. and now you can see that top simplifies to negative three t negative three and then the bottom is six minus three t square root plus 3 and what you can do you can pull 3 out and also you can cancel the negative sign you can write this on as 3 t plus 1 over 6 plus i know six plus six minus six minus three t square root plus three ah and then what we can do is we can use a similar argument and you can see that this is uh less than this is less than uh or equal because it can be zero so less than or equal to you could write this as t plus one times three from the top and you can see the largest value that means you're going to pick the smallest value from the bottom which is three so you can write it like that and then what we can do is uh we're gonna pick a delta which is smaller and the same argument so we can see that so this simplifies to simply 1 3 3 means 1 so that means we pick a delta we pick a delta so that this is smaller than um this less than or equal to epsilon so again just to make sure that you do not get any undefined values as before are we're going to restrict the domain and we take epsilon in the minimum between 1 and epsilon you can see this argument is very clear now so so we can see taking taking a delta which is the minimum of 1 and epsilon we have we have 0 less than rt minus minus 1. so you can write it like that that's what t plus one means this then delta implies that um t plus one it's the same thing less than epsilon because the minimum tilde is the minimum of the two then it should be smaller than any of them now um again we mult we can write this one as this exactly same as te t plus 1 over so 6 minus 3 t because this is a smaller value plus 3 we multiply by something smaller will be the largest possibility so this is smaller and then you can see that this simplifies 2 because of the equality 6 minus 3t let's say you have to do that first 6 minus 3t a minus 3 absolute value less than epsilon so that's exactly what and you can see that this implication works very well so then we can see this somewhere the compute does for all epsilon uh there is a delta positive such that for all x in the domain of the function that's very important the function zero because some uh sometimes we don't write this one because that gonna make the whole argument wrong because if you can use if you get some value which are not in the domain the implication does not follow so it is x so t minus minus 1 less than delta implies t and six minus three t square root minus three less than epsilon so that's exactly the epsilon delta argument so therefore you can see therefore by the definition followed by the definition limit t goes to negative one six minus three t equal three so that completes the proof so let's do one last problem number eight so as the next problem we have limit x goes to five seven over x minus one square root which is which is equal to seven half so this is like a rational and root so let's see how the algebra works so the proof so we're going to say let epsilon positive then note that but we're going to begin with the algebra first so we write 7 over x minus 1 square root minus 7 half equal this is equal so we can pull 7 then we're going to get 1 over x minus 1 square root 1 half now we can do the cross multiplication and write this as 2 minus x minus 1 square root of 2 times x minus 1 square root ah so what what happened here is when you have a b minus c d you can write this one as a d minus b c over b d so this is how we normally do this algebra so that's the cross multiplication you can see the a d minus b c the cross multiplication you can always do that ah so now what we can do is since we know that x minus 1 square root is greater than or equal to 0 we're going to pull that out of the absolute sign so that means we can write this as 7 over 2 x minus 1 square root and then the rest we have 2 minus x minus 1 square root so what we're going to do now this is just so like square root type of problem that means we're going to use the conjugate method now so which is 7 over 2 x minus 1 square root so we're going to multiply the top number term by square root so conjugate so it is x minus 1 square root and be careful conjugate has changed the sign only of this one not inside so that means times the conjugate would be 2 plus x minus 1 square root then 2 plus x minus 1 square because this is simply 1. so now you can simplify the top using the difference of squares argument a minus p a plus b equal a squared minus b squared that means you can square the top so we're going to get 7 over 2 x minus 1 square root so you can see the top is 4 minus you square x minus 1 at the top and then the bottom you can see it is 2 plus x minus 1 square root that you can bring out because it's always positive so that means now what we can do is you can see this simplifies very well so we can write this one as uh the top is just 5 minus x which you can write as x minus 5 so you can say x minus 5 absolute value that's a top that is this one then i'm going to combine everything else to one expression so the rest you can write as 7 over 2 x minus 1 square root and the other whole expression you can write it because it's positive so you don't need the absolute sign anymore you can write this one as x minus 1 yep write it like that now what we're gonna do we're gonna find the largest value of this expression uh so that's not that difficult to find we're gonna do a similar argument we restrict the domain so once you receive the domain so what we can do we're gonna say okay we're gonna pick values x minus five less than one again it still works sound of this x equal the sort of this one may not work then you might have to have for something smaller one because you want to make sure that you do not include any like you in this case you do not include x equal one so that's the troublemaker so x cannot be one and also actually you can see uh so you can have so what it says that uh uh what it says that because this is square root you want to make sure that x is bigger than one because you can't have any smaller value you cannot have zero there so it's actually bigger than one so that's what you make sure that and then if you do it like that it's exactly work because you're gonna see five around like you go only one away so that means it's going to be 6 and 4 so everything looks good so now since all these terms are in the denominator so what we're trying to do is uh we're going to pick smaller values so we're going to pick smaller values so you can be the smallest value when you apply the smallest value you get the largest fraction and you can see that if you plug in values for example if you plug in 4 the x minus 1 square root is going to be square root 3 so the square root 3 so that means you can see that this this expression is you can see so this is smaller than so you can see that this expression in red is smaller than 2 because the bottom can be larger than larger than four since the bottom can be larger than four since expression can be um the largest value is smaller than two so that means what we can do we're gonna find a delta so that two times delta the 2 is much bigger than this fraction uh less than so you find a smaller delta so that 2 times delta is smaller than epsilon now we can use this to write the argument so so you can see uh so then uh choose so you can see choosing delta as the minimum of one and epsilon over so you can see two because we have two delta so what we have is two delta uh less than less than equal epsilon that means delta is less than equal epsilon over 2 so then we can write so we have so we have all the nice implications of 0 less than x minus 5 less than delta it implies uh x minus five are less than epsilon over two because you can be anything you like then multiply by two we can write this one as x minus 5 times 2 less than epsilon since 2 is the largest value you can replace this one by something smaller so you can write this as 7 over the large expression 2 x minus 1 square root times 2 plus x minus 1 square root less than epsilon and now you can see that this all simplifies to a 7 over x minus one but that's what we get over seven half minus seven half less than epsilon so therefore so we have you can see therefore by the definition by the definition we prove that the limit x goes to 5 7 over x minus 1 square root equals 7 half so this completes the proof so you can try the last one as an exercise so see whether you can prove this problem as an exercise this will be difficult thank you for watching you

in this video we solve several limit problems using the epsilon delta definition uh you may watch this video after watching the oge video that discuss one problem and also the theory video that explain the theory behind the epsilon delta definition you can find the link in the description so in this video are we gonna solve uh several problems in different ranges for example we have linear problems and quadratic and quadratic with the two and three terms and cubic and square root functions uh rational functions and then also square root in rational form so these are all type of problems we have nine problems here so let&#39;s see how we solve each one so i&#39;m going to use a different logical arguments like uh sometimes double implicit implications and sometimes single implicit implications and also sometimes like a simple calculations so let&#39;s see how we can solve each one so let&#39;s start with the uh first let&#39;s start with the definition so the precise definition so what&#39;s a precise definition so we say that the limit when x goes to a f x equal l are if and only if but because a double implication if and only if for all epsilon that means for any given epsilon there is a delta which is a function of epsilon such that for all x in the domain of the function because this is very important especially when there is a discontinuity so for squared functions or rational function you have to be very careful with this one ah then t 0 less than x minus a less than delta implies uh the effects the difference between the function and the limit is less than epsilon so if you can prove that for all epsilon there&#39;s a delta and for all x in the domain uh the first one implies the second one that means if x is within delta this distance then you want to make sure that the function is within the l uh the functions function is within the epsilon distance to the l so so that&#39;s the what we call the precise diffusion or the limit definition good so uh so let&#39;s look at this problem so we&#39;re going to go to the first problem so what it says that so we have um so prove that prove that the limit when x goes to uh let&#39;s say eight just pick a number a one half x minus three since you&#39;re more terms you&#39;re able to parent this around equal one that&#39;s what we&#39;re trying to do we have this one coming from and you can see that one half x minus 3 is a linear function this is a polynomial so that means you can just plug in the value to find because these are continuous functions so you can plug in the value you plug in eight and you can see that you get one so that&#39;s where the one coming from but how we so what we&#39;re trying to do right now is we&#39;re gonna prove that that&#39;s exactly the result so uh so to do uh so to prove a problem like this what we&#39;re going to do so we can use the epsilon definition so we pick a general epsilon so any epsilon and then we try to find a delta that implies uh this condition so uh so let&#39;s see how the argument gonna work so the proof so what we can do we&#39;re gonna pick a uh epsilon let&#39;s say let epsilon positive so you pick a positive epsilon and then what we need to do so need to show that so let&#39;s see how to write it like a logical complete logical proof okay so i need to show that need to show that there is a delta there is a there is a delta which is a function of epsilon let me depend on epsilon such that such that 0 x minus 8 less than delta implies the function the function is one half x minus three minus the limit in this case it is one actually minus one that&#39;s the element we are talking about 1 is less than epsilon so the goal is to find the delta so and then you can see that this is the function is the function and that is l so this is uh l so what we normally do is uh because we don&#39;t know what the delta is so we we start with what we need to prove and then think about the condition and go forward so what we do so we can see now are we gonna pick what we want to prove so what you want to prove is one half x minus three minus one less than epsilon that&#39;s what we want to prove so and then we we know that this if and only if this double implication we&#39;re going to simplify inside so if you simplify inside what you get one half x minus 4 which is less than epsilon so they are they&#39;re equivalent now what we&#39;re going to do we&#39;re going to multiply both sides by one half so that means this is if and only if you multiply inside like both sides you can just uh bring inside by uh by 2. why we do that your goal is to create the x minus eight term now when you multiply you can see this is again if and only if x minus or you&#39;re gonna get x minus eight less than a two epsilon so you can see that this is exactly what we want to prove so what we&#39;re going to do so this can give an option so you can see that if you pick this part as delta then this implies that i implies uh the condition so that means this simply gonna give you uh the delta so one good example uh option would be delta equal to epsilon you really don&#39;t want to do that because you can pick any delta which is smaller than this so one what do you say optimal option would be here it&#39;s not the optimal option so would be here two epsilon but you can pick any delta which is smaller than two epsilon for example you can pick delta equal epsilon that still works because because delta is the uh uh the gap or the radius uh from eight so that means even if you pick a smaller one that still works but that&#39;s still within that so that means you can pick any smaller um delta for this one but we&#39;re gonna go with this one because that&#39;s what we directly get from this one but again you can pick anything so now we can say so thus so it&#39;s very simple now argument does because it&#39;s only final leaf that means if you start with one you get the other one so that&#39;s a choosing just choosing delta equal to epsilon or you can pick any for example you can pick the like epsilon that we can do later in another example so we have we have because of this all double implications you can say 0 less than x minus 8 are less than delta implies now we can write just all like one direction for example you can write this one go this way and then this one go that way and all go like a backward only only like a left so then you can you get the answer so this implies that uh the first one so you can say one half x minus three minus l in this case the 1 are less than epsilon that&#39;s exactly what you want to prove so there&#39;s a function and this is the limit so this completes the proof this is the epsilon delta argument so this complete because if you pick any epsilon and you prove that the iso delta that works okay so this completes the proof this completes the proof good so you can see that&#39;s very simple it&#39;s not that difficult but there are some logical arguments that you want to make so let&#39;s look at the next problem so as the next one we have proved that limit x goes to 2 x cube equal eight again uh x cube bc polynomial so you can plug in the value and get the uh limit so which is eight but let&#39;s see how we can prove this so in this time i&#39;m gonna use a slightly different argument and let&#39;s see how it works so the proof proof let&#39;s say let epsilon be positive so you&#39;re picking a positive epsilon ah then note that note that uh so we can start what we want to prove we&#39;re going to start with x 3 minus 8 equal and we know that we can factor this how to factor this so you can see the 8 means 2 to the 3. so we you can use the difference of cube formula we know that a3 minus b3 equal a minus b the first factor the second factor is a squared plus a b plus b squared so that is how you&#39;re going to factor cubes since uh 8 is 2 q we can use that so because with that if you look at those two factors we can write this like this the absolute value the first factor minus times the second factor you can add it like that because it&#39;s a product plus 2x plus so this is true they are equal now what we&#39;re going to do so the goal is to find the log because x minus 2 that so we need x minus 2 that&#39;s we already have so the goal is to find the largest value of this factor so find the largest value of the rest of the function find the rest of the expression find the largest value that&#39;s how we need largest ready find the largest value but the problem is we don&#39;t know what&#39;s the largest value because this is just a polynomial it can have any any value because it goes to infinity so there&#39;s no largest value here so that means we need to restrict uh the domain so what we&#39;re gonna do we&#39;re gonna restrict x minus two so we&#39;re gonna say um so we can start with like uh so um so let x minus 2 is within 1. so we&#39;re going to pick values you can pick any value you like so let&#39;s say 1. so that means what that means if you look at this interval what this means it means that the all the values which are one unit away from two so that means uh you add once you get three and you subtract you get one so those are all the values that you&#39;re gonna get since we are looking for the largest value i&#39;m going to put the largest value here so pick the largest value if you&#39;re the largest cell in this case 3 is it&#39;s not in there but we are looking for a bound so that means you can say 3 here what we can do we&#39;re going to see what will happen this this that factor so if you put x equal 3 x squared means 9 plus 2 times 3 means 6 plus 4 that means this simplifies to 19 so this you can see that the largest possible value under this condition that you can get is 19 so that means uh but if you restrict uh this one to delta so if you restrict this sort of delta let&#39;s say you this is smaller than delta that means the largest you can get here is 19 delta because the other one gives you 90. so what your goal is to find the largest value of the other factor so the condition is you want to make sure that this is smaller than epsilon so what we&#39;re going to do we&#39;re going to pick a really smaller delta such that 19 times delta you don&#39;t need 90. you can say 20 here for the safe side but 19 is the largest value you can get so you what you can do you&#39;re going to pick a really small delta so that the 19 delta is smaller than epsilon so you can always do that you can pick a really really small delta that means you can make sure that 19 delta is smaller than epsilon actually you can even have equal signs so we already have a strict inequality so that&#39;s that&#39;s exactly how you&#39;re going to find delta so after you get that you want to make sure that you&#39;re going to pick a really smaller delta such that 19 delta is smaller than or equal to epsilon so that means now we&#39;re going to solve this so if you solve this you can see this is 19 uh delta smaller than epsilon this is the delta is less than or equal to is epsilon over 19. you can pick any like for example you can pick epsilon or 20. so anything works so it&#39;s 19 or more now we have all the conditions we need so that means now what happened so so uh we have now and so the condition is and so this says that x minus ah 2 less than epsilon over 19. now what we&#39;re going to do you want to make sure that both conditions are satisfied so we&#39;re going to pick pick a delta which is the smallest of the two what are we going to do so so what we&#39;re gonna now choose so this interesting part now choose delta which is the minimum of those two minimum of 1 or and epsilon 90 because if you pick the minimum then it&#39;s guaranteed that it satisfy both of them so now let&#39;s see how the rest of the argument works so then so you can write so you can write this then you can say if epsilon 0 less than x minus in this case 8 actually uh it is 2 so 2 less than delta this implies that because since delta is the minimum this implies that x minus 2 less than any of them so that means for example epsilon 90 will be the minimum since we pick the minimum it is smaller than any of them now but we&#39;re going to multiply by 19 both sides so it&#39;s x minus 2 times 19 less than epsilon but we know that since 90 is the largest value we can bring that back a factor back here so we can write we can bring the other factor here so it is x squared plus 2x plus 4 that&#39;s the other factor which is less than epsilon but now we can combine them and we know that this is this side is smaller than 19 so this implication works so this implication is true because 90 is the largest value so if you replace something by something smaller then still it should be valid now we&#39;re going to combine them you&#39;re going to multiply through once you multiply through so you could write this one i would write the extra step here you really don&#39;t need that and you can bring into one product like that now you can see that because of the uh difference of cubes argument this simply equals to x 3 minus 8 less than epsilon and that&#39;s exactly what we will produce that&#39;s exactly how the argument works so it&#39;s directly implies uh that way so that means so thus what we prove so we prove that for all epsilon there is a delta which is a which is depend on epsilon such that 0 x minus 2 less than delta implies x 3 minus 8 less than epsilon that&#39;s exactly the definition so therefore by the definition let&#39;s have you write the argument therefore by the definition therefore by the definition the limit x goes to 2 x cubed equal 8. so this completes the proof so you can see this is a really nice neat argument so let&#39;s look at the next example so as the next example we have proved that the limit when t goes to three t plus one square root equal to um so we&#39;re gonna we prove and we we notice something here we see that this is called function we wanna make sure this part is greater than or equal to 0 so this implies t greater than equal negative 1 so that means you want to make sure that you are not going to plug in anything uh which is smaller than negative 1 let&#39;s see how we can take care of that later so again uh so we&#39;re gonna use a similar argument so we&#39;re gonna say let ah epsilon greater than zero uh then note that know that so what we&#39;re going to do we&#39;re going to start with what we need to prove note that t plus 1 square root minus 2 that&#39;s the one we&#39;re going to consider subtract the limit from the function so what we&#39;re going to we&#39;re going to simplify this we are not going to write anything with epsilon here we just simplify this expression so this is equals what we&#39;re going to do when you get square root what we normally do is we&#39;re going to multiply by a special one where the special one is the conjugate top time bottom the numerator and the denominator so that&#39;s the argument we&#39;re going to use here so that means we have t plus 1 square root minus 2 times you&#39;re going to multiply the top and bottom by the conjugate conjugate means you change the sign here in between not inside change the sign for the subtraction so uh so go plus sign and then also we do that with the minus sign so so this is simply 1 so you can always multiply by 1 so we do that and it&#39;s under ah absolute sign now you can see that when you uh multiply the top we can use this result we know that if you have a minus b with a plus b you&#39;re gonna get a squared minus b squared that&#39;s exactly the reason why we multiply by the conjugate so so one is the conjugate of the other so and because of that you&#39;re going to square the first one so when you square the first one you&#39;ll get t plus one minus minor second one you&#39;re going to get 4 so that&#39;s what you get for the top for the bottom and you can see that this is always greater than or equal to 0 x is greater than or equal to 1 um this is square root so it&#39;s always we assume that&#39;s greater than equal to 0 so because of that this whole expression is positive so that means you can so we still have the absolute value for the top but for the bottom you can take it out because that&#39;s a positive term so plus 2 and then so that&#39;s what you get now uh you can see this again simplifies to the top you know simplifies to t minus 3 so that&#39;s exactly the term that we want to bound and the other thing i&#39;m going to write as a separate term uh because then the argument would be much easier so we have this so now what we&#39;re going to do is we&#39;re going to make you want to find the largest value of this expression you&#39;ll find the largest value of this expression but this one is very clear because we know that the t plus one square root is greater than or equal to zero so that means the largest value you get when this is zero just that&#39;s when you get the largest value uh so that means you can see that this whole expression is less than or equal to one-half so you can write it like that so that means you can write this on s so this is less than or equal to you can see this is less than or equal to uh but if you if you bound uh this one by delta because this is smaller you can see that this is smaller than one-half delta one half coming from uh the other expression we know that this expression the whole expression is less than equal one half one half you get uh the t when t equals negative one that&#39;s when you get uh one half otherwise this is smaller because you are dividing the numerator by something so that means this is smaller so always positive that is we add a positive term to the new denominator that means the fraction gets smaller so um and then and then uh so what we&#39;re going to do that we&#39;re going to pick a small delta so that this one-half delta is smaller than epsilon that&#39;s what we&#39;re trying to do so we&#39;re going to pick a delta smaller delta which is uh so that one-half delta is smaller than epsilon so that means if you solve for delta you can see that delta is smaller so what you get from here you see that delta is smaller than ah less than equal to 2 epsilon so that means you can pick either two epsilon or any other epsilon uh any other uh function delta which is smaller than uh two epsilon for example delta equal epsilon but uh but there&#39;s something you have to be careful because you know that you want to make sure that you do not pick any values of t which are less than uh negative 1. to guarantee that what we&#39;re going to do we&#39;re going to put extra condition on delta so you say that so now what we&#39;re going to do now choose this interesting part now choose now choose delta equal the minimum of just put a number here like one or one half something like that so that it doesn&#39;t go until ah negative one so you can see that if you uh put a delta like that uh so you can see what we&#39;re trying to do is we try to go this is three so you start to stay around this interval four and two you can see that the negative one is is far away so that means it&#39;s never reached negative one that means there is no any discontinuity here like there are no any undefined point this is exactly what how to do that we just put a um some number so that doesn&#39;t go that far that&#39;s why i put one here if you don&#39;t put that there the argument wrong and then we have and then we get the minimum of 1 and 2 epsilon so once you do that then you can write a 0 less than a t minus 3 less than delta implies that um t minus 3 less than any of them so for example 2 epsilon because the minimum will repeat the delta as the meaning of c is the minimum it is smaller than any of this more than one also smaller than the other one two epsilon now what we can do we&#39;re gonna divide both sides by two so we can write this one as t minus three times one half dividing both sides by 2. now what we can do we know that one-half is the largest value of the expression so that means we can put that back in um so we&#39;re going to put that back in so we say t minus 3 times we&#39;re going to put the expression here so what is the expression it is t plus 1 square root plus 2 less than epsilon why because we know that that expression is bigger than a one-half one have the largest values if you put something smaller it is small uh now what we can do are we gonna write backward and we know that this is if and only if uh you put the multiplied by the conjugate so this is fine only so they are equal so what we can do t minus 3 we can write we can break it into t plus 1 square root minus 2 times t plus 1 square root plus 2. so the benefit is because we know that this is exactly steps that is that&#39;s why we can write this like otherwise it&#39;s going to be like very surprising where is coming from but we know that this is exactly how we simplify this uh so now what we&#39;re going to do we&#39;re going to combine uh you can see you can cancel this with that so once you cancel that so this is again different only because they&#39;re all equal so if an only if you can write this one as t plus 1 square root minus 2 less than epsilon so now we have the proofs that&#39;s exactly what we want to prove so therefore you can say therefore by the definition by the definition epsilon delta diffuse by the definition you prove that the limit t goes to r r3 t plus 1 square root equal to so that completes the proof so let&#39;s look at the next problem so we have uh prove that limit x goes to 2 3 x square plus 4 equals 16 so we&#39;re going to use a similar argument so the proof uh so let epsilon be positive and then say consider consider so we&#39;re going to consider what we need to prove sorry 3x squared plus 4 a minus 16. but we know that this is equals to if you simplify you can see that you&#39;re gonna get 3x squared minus 12 that&#39;s what you get and then what we can do is we can factor we can pull 3 out so if you pull 3 out as a factor you&#39;re going to get x squared minus 4. now we can factor x square minus 4 using a different squares so that means what we can do is we can write the x minus 2 here and we separate the rest of the rest from this one so we do the multiplication three times x plus two and then you can also combine and you can write it like this so we&#39;re going to say x minus 2 you&#39;re going to combine it and we&#39;re going to put 3 x plus 6 we&#39;re going to combine like that and now the goal is to abound the second term so to do that we do as we because we know that this there&#39;s no uh otherwise there&#39;s no largest value of that expression so we define the largest value of this expression but there&#39;s no largest value it goes infinity so what we&#39;re going to begin about the delta so what we do as before we&#39;re going to say are we going to pick a value x minus 2 less than 1 because that&#39;s going to rescue the x value so this says that x is within 1 that means largest value is 3 and the smallest value is 1. since we are looking for the largest value uh we&#39;re gonna pick so that&#39;s the largest value three so we&#39;re going to use that value uh so that means if you use three and you can see that it&#39;s a three times x uh so it&#39;s 3 times 3 plus 9 that means this is less than 19 because this is an open interval so that means what we what we will make sure that is so we will make sure that this is less than epsilon so what we&#39;re going to do we&#39;re going to find a delta we&#39;re going to find a delta which is smaller than in this case actually you get 15 so it&#39;s not 90 50. so um so we&#39;re gonna we&#39;re gonna find a delta so that the 15 times delta is smaller than epsilon uh because we make sure that this is less than delta so that&#39;s where the 15 coming from so 15 times delta we make sure that we finally a smaller delta so that 15 delta is smaller than epsilon that you can always do so pick a delta like that um so now we can write the rest of the argument so now we can say now choose so now i chose a delta which is the minimum of uh it&#39;s a minimum of one and so you can see when you simplify for delta you&#39;re gonna get uh delta is less than epsilon over 15. so you can pick this or any smaller values so uh so we got epsilon of 50. so you can say 720 that works so then what happened then because of this equality and you can see that then 0 less than x minus in this case 2 less than delta implies that x minus 2 times x minus 2 absolute x 1 is 2 less than we can put epsilon over 15 so this implies x minus 2 times 15 less than epsilon now what we&#39;re going to do we&#39;re gonna replace 15 by the expression because if the expression says that is smaller the implication works so what we do instead of 15 now we can write 3x plus 6 less than epsilon that implies that that directly implies that what we want since the um the left hand side is equals to this one so you can write this one as 3 x squared plus 4 minus 16 less than epsilon because they are all equal so that completes the proof this completes the proof so that&#39;s how you make the a logical argument for for a problem like this so let&#39;s look at the next one so as the next one we have limit x goes to 1 4 plus x minus 3x squared equal to so we do exactly the same way we did before so the proof so we&#39;re going to say let epsilon are positive then what we can do is we can write several implications so we can start with what we want to prove so it is for plus x minus 3 x squared a minus 2 which is less than epsilon that&#39;s having proof this implies like it&#39;s like double implication this if and only if uh but we&#39;re gonna we&#39;re gonna simplify the left hand side so if you simplify you&#39;re gonna get 2 plus x minus 3 x squared are less than epsilon this is if and only if you know that inside the absolutely you can switch the order so we can write this one as 3x squared so you can simply multiply by negative sign so 3x squared and then minus x minus 2 that&#39;s exactly the same you can only switch inside the absolute sign now we can factor so you can see that this says factors is again if an only if they are all equal x minus 1 times the other factor is 3x plus 2 less than epsilon now this is again if and only if you can write this one as if you pull the x minus 1 out or you get the other factor which is 3x plus 2 3x uh plus 2. so now what we want to do is to find the largest value of this one so this is very similar to the previous one what we&#39;re going to be going to bound x minus 1 first so we can say that so we&#39;re gonna bound that so we use the same trick every time so this is around one so we&#39;re gonna go uh we&#39;re gonna pick x minus one uh less than one again so that&#39;s what we&#39;re trying to do so that means it&#39;s going to go up to 2 and 0. so 0 and 2 since we are looking for the largest value we pick this side because we pick 2 here so that&#39;s the largest value so we&#39;re going to pick that one so that means then we can see that uh if we the largest values x equal 2 so you&#39;ll get 3x plus 2 equals 8. so that means now we have only single implication it&#39;s only go left side so this you can see that then we can write if you pick the largest value we have implications to the left side so it&#39;s gonna see repeat the larger that&#39;s the largest value of three x plus two uh so it&#39;s eight uh less than so what we can do we wanna make sure that now we&#39;re going to pick a delta so that 8 times delta is smaller than epsilon so that means we&#39;re going to pick a we can now we&#39;re going to pick a smaller delta so you look at this smaller delta so that 8 times so that 8 times delta so it doesn&#39;t both looks like 8 so it is 8 times delta is less than or equal to epsilon that means if you simplify for delta you&#39;re going to see that delta is are less than equal epsilon over eight so that means then what we can do we can pick a value that satisfy both conditions so that means now we say choose choose delta equal the minimum of the two minimum of one and epsilon number eight so then we have this uh simplification so we can see that because of this double implication we can simply write 0 less than x minus 1 less than delta implies because the double implication we can say x minus 1 less than epsilon over eight so complicating because that&#39;s the smallest now um this implies that uh four plus x minus three x squared minus 2 less than epsilon so that completes the proof so as the next one we have limit i u goes to 1 3 over u plus 2 equal 1 and you can say this is a rational function so you want to make sure that you do not include the value uh negative two with that value gonna give you something undefined so uh let&#39;s see how it works so the proof uh so let epsilon be positive then you can write this statement you can say a 3 over u plus 2 minus 1 that&#39;s the difference between the functional limit less than epsilon if and only we&#39;re going to simplify inside so if you simplify inside that we&#39;re going to do cross multiplication you can write this on as 3 minus u plus two over r u plus two uh less than epsilon so this is final leaf and again this is phenol leaf you can see at the top you can simplify as 1 minus u absolute value and then again u plus 2 absolutely since we don&#39;t know what is positive so we&#39;re going to leave it like that and then you can also switch the top because it&#39;s on the absolute sign so you can switch the top and also isolate this term u plus 2 absolute value now your goal is to find the largest value of this expression that&#39;s not that difficult so what we can do as before we can restrict the domain for u so that means so let&#39;s do that uh so u is around 1 and then again so we&#39;re gonna say u minus 1 are less than 1 not x so u minus 1 are less than 1 so when you do that then you&#39;re going to see that this is 2 and 0. so now so let&#39;s see that so we&#39;re going to pick the values in this interval so we&#39;re going to pick the values only in this interval but we are looking for the largest possible value since this is in the denominator now we&#39;re going to pick the value on the other corner because this is the value we&#39;re going to pick so now we pick the smallest value smallest 20. it will be the smallest value you get the largest fraction so once you plug in the smallest value that means 0 and you can see that you can write this sentence so we can see so this is uh left implied that u minus 1 absolute value now you&#39;re going to pick the largest value with strip with the largest value then it&#39;s going to give you a left implication so ah less than now again we restrict this side u minus 1 by delta so that means uh you can see that we want to make sure that this is within the less than epsilon so what we can do we&#39;re going to pick a value a delta so that the half times delta we have is the largest possible value you can get from the other expression so ah the half delta so you pick a smaller delta so that half delta is smaller than epsilon so it&#39;s very clear so that means uh and then we&#39;re gonna we&#39;re gonna use this to simplify for delta so this is the delta is smaller than two epsilon so we&#39;re gonna choose a delta which is smaller than one and two epsilon so now uh so it&#39;s as before so you can see that uh choose so i&#39;m going to show you a different argument this time so let&#39;s say choose because you know that you can pick either a delta equal arms it&#39;s a less than or equal so it&#39;s a delta equal uh 2 epsilon or any any smaller one so let&#39;s say for example you&#39;re going to pick delta equal epsilon this time and see how the argument works so delta equal minimum so you can pick either minimum of 1 and 2 epsilon or even any other estimator so let&#39;s let&#39;s say you pick epsilon in this case so let&#39;s see how the argument works now so then you can write this then 0 less than u minus 1 less than delta implies that u minus 1 is less than epsilon because that&#39;s how we pick the delta which has to be smaller than uh any of them but you know that this is smaller than 2 epsilon so that&#39;s how you&#39;re going to jump through this step now what we can do we&#39;re going to skip the epsilon we&#39;re going to go with the 2 epsilon so that means this implies now you can say that u minus 1 uh i&#39;m going to divide by half so i&#39;m going to say okay let&#39;s try this step so this is true now now what we can do we can divide by 2 both sides and then since epsilon one half is the largest value this implies u minus one so we&#39;re going to put u uh plus two absolute value back in there now you&#39;re going to combine so you can see that because of the argument we write because the double implication we wrote this is exactly similar to 3 over u plus 2 minus 1 that&#39;s exactly similar less than epsilon so that means that&#39;s exactly what we want to prove so this completes the proof this completes the proof so you can see that it doesn&#39;t take much time and if you understand the how the argument works and so you can see that we can even pick a small epsilon but still the argument works good so let&#39;s look at the next one so as the next one we have limit t goes to negative 1 6 minus 3 d square root equal 3 um so we&#39;re going to do a similar argument here but we&#39;re going to use the simplification step now no no uh implications so let&#39;s say proof so we pick a epsilon so let epsilon be positive so p capacity epsilon and note that so i&#39;ll write different type of proofs so that you can learn different arguments so let&#39;s say we can start with what we want to prove 6 minus 3 t square root minus 3 that&#39;s the limit and the function this is equals to again we&#39;re gonna multiply the uh top and bottom by the so we have six minus three t square root minus 3 that&#39;s the first term times we&#39;re going to multiply the conjugate of this so we only change the sign between not inside so that means we have 6 minus three t square root plus three and we divide by the same thing six minus three t square root plus three uh no epsilon here now what we can do we&#39;re going to combine the top so it&#39;s because of the differential squares formula you can see the top is now 6 minus 3t minus 9 and we would absolutely sign there and then the bottom bottom you can see 6 minus 3t square root plus 3. we don&#39;t need the absolute sign the reason is this term is greater than equals 0. and now you can see that top simplifies to negative three t negative three and then the bottom is six minus three t square root plus 3 and what you can do you can pull 3 out and also you can cancel the negative sign you can write this on as 3 t plus 1 over 6 plus i know six plus six minus six minus three t square root plus three ah and then what we can do is we can use a similar argument and you can see that this is uh less than this is less than uh or equal because it can be zero so less than or equal to you could write this as t plus one times three from the top and you can see the largest value that means you&#39;re going to pick the smallest value from the bottom which is three so you can write it like that and then what we can do is uh we&#39;re gonna pick a delta which is smaller and the same argument so we can see that so this simplifies to simply 1 3 3 means 1 so that means we pick a delta we pick a delta so that this is smaller than um this less than or equal to epsilon so again just to make sure that you do not get any undefined values as before are we&#39;re going to restrict the domain and we take epsilon in the minimum between 1 and epsilon you can see this argument is very clear now so so we can see taking taking a delta which is the minimum of 1 and epsilon we have we have 0 less than rt minus minus 1. so you can write it like that that&#39;s what t plus one means this then delta implies that um t plus one it&#39;s the same thing less than epsilon because the minimum tilde is the minimum of the two then it should be smaller than any of them now um again we mult we can write this one as this exactly same as te t plus 1 over so 6 minus 3 t because this is a smaller value plus 3 we multiply by something smaller will be the largest possibility so this is smaller and then you can see that this simplifies 2 because of the equality 6 minus 3t let&#39;s say you have to do that first 6 minus 3t a minus 3 absolute value less than epsilon so that&#39;s exactly what and you can see that this implication works very well so then we can see this somewhere the compute does for all epsilon uh there is a delta positive such that for all x in the domain of the function that&#39;s very important the function zero because some uh sometimes we don&#39;t write this one because that gonna make the whole argument wrong because if you can use if you get some value which are not in the domain the implication does not follow so it is x so t minus minus 1 less than delta implies t and six minus three t square root minus three less than epsilon so that&#39;s exactly the epsilon delta argument so therefore you can see therefore by the definition followed by the definition limit t goes to negative one six minus three t equal three so that completes the proof so let&#39;s do one last problem number eight so as the next problem we have limit x goes to five seven over x minus one square root which is which is equal to seven half so this is like a rational and root so let&#39;s see how the algebra works so the proof so we&#39;re going to say let epsilon positive then note that but we&#39;re going to begin with the algebra first so we write 7 over x minus 1 square root minus 7 half equal this is equal so we can pull 7 then we&#39;re going to get 1 over x minus 1 square root 1 half now we can do the cross multiplication and write this as 2 minus x minus 1 square root of 2 times x minus 1 square root ah so what what happened here is when you have a b minus c d you can write this one as a d minus b c over b d so this is how we normally do this algebra so that&#39;s the cross multiplication you can see the a d minus b c the cross multiplication you can always do that ah so now what we can do is since we know that x minus 1 square root is greater than or equal to 0 we&#39;re going to pull that out of the absolute sign so that means we can write this as 7 over 2 x minus 1 square root and then the rest we have 2 minus x minus 1 square root so what we&#39;re going to do now this is just so like square root type of problem that means we&#39;re going to use the conjugate method now so which is 7 over 2 x minus 1 square root so we&#39;re going to multiply the top number term by square root so conjugate so it is x minus 1 square root and be careful conjugate has changed the sign only of this one not inside so that means times the conjugate would be 2 plus x minus 1 square root then 2 plus x minus 1 square because this is simply 1. so now you can simplify the top using the difference of squares argument a minus p a plus b equal a squared minus b squared that means you can square the top so we&#39;re going to get 7 over 2 x minus 1 square root so you can see the top is 4 minus you square x minus 1 at the top and then the bottom you can see it is 2 plus x minus 1 square root that you can bring out because it&#39;s always positive so that means now what we can do is you can see this simplifies very well so we can write this one as uh the top is just 5 minus x which you can write as x minus 5 so you can say x minus 5 absolute value that&#39;s a top that is this one then i&#39;m going to combine everything else to one expression so the rest you can write as 7 over 2 x minus 1 square root and the other whole expression you can write it because it&#39;s positive so you don&#39;t need the absolute sign anymore you can write this one as x minus 1 yep write it like that now what we&#39;re gonna do we&#39;re gonna find the largest value of this expression uh so that&#39;s not that difficult to find we&#39;re gonna do a similar argument we restrict the domain so once you receive the domain so what we can do we&#39;re gonna say okay we&#39;re gonna pick values x minus five less than one again it still works sound of this x equal the sort of this one may not work then you might have to have for something smaller one because you want to make sure that you do not include any like you in this case you do not include x equal one so that&#39;s the troublemaker so x cannot be one and also actually you can see uh so you can have so what it says that uh uh what it says that because this is square root you want to make sure that x is bigger than one because you can&#39;t have any smaller value you cannot have zero there so it&#39;s actually bigger than one so that&#39;s what you make sure that and then if you do it like that it&#39;s exactly work because you&#39;re gonna see five around like you go only one away so that means it&#39;s going to be 6 and 4 so everything looks good so now since all these terms are in the denominator so what we&#39;re trying to do is uh we&#39;re going to pick smaller values so we&#39;re going to pick smaller values so you can be the smallest value when you apply the smallest value you get the largest fraction and you can see that if you plug in values for example if you plug in 4 the x minus 1 square root is going to be square root 3 so the square root 3 so that means you can see that this this expression is you can see so this is smaller than so you can see that this expression in red is smaller than 2 because the bottom can be larger than larger than four since the bottom can be larger than four since expression can be um the largest value is smaller than two so that means what we can do we&#39;re gonna find a delta so that two times delta the 2 is much bigger than this fraction uh less than so you find a smaller delta so that 2 times delta is smaller than epsilon now we can use this to write the argument so so you can see uh so then uh choose so you can see choosing delta as the minimum of one and epsilon over so you can see two because we have two delta so what we have is two delta uh less than less than equal epsilon that means delta is less than equal epsilon over 2 so then we can write so we have so we have all the nice implications of 0 less than x minus 5 less than delta it implies uh x minus five are less than epsilon over two because you can be anything you like then multiply by two we can write this one as x minus 5 times 2 less than epsilon since 2 is the largest value you can replace this one by something smaller so you can write this as 7 over the large expression 2 x minus 1 square root times 2 plus x minus 1 square root less than epsilon and now you can see that this all simplifies to a 7 over x minus one but that&#39;s what we get over seven half minus seven half less than epsilon so therefore so we have you can see therefore by the definition by the definition we prove that the limit x goes to 5 7 over x minus 1 square root equals 7 half so this completes the proof so you can try the last one as an exercise so see whether you can prove this problem as an exercise this will be difficult thank you for watching you

Transcript for:Epsilon-Delta Definition of Limits: Solving Problems

Transcript for:
Epsilon-Delta Definition of Limits: Solving Problems