in this video I'm going to be working some limit examples for kind of at the beginning of AP Calculus so really at this point we really only have the definition of a limit that you know a limit exists if each of the one-sided limits exist and degree so we're going to start with this one and we're going to find the value of the limit of f ofx as X approaches one or explain why it doesn't exist so in order to do that we're going to look at the two one-sided limits so I'm going to start with this cosine of pix when X is greater than or equal to 1 well if x is greater than or equal to 1 that kind of corresponds to the Limit as X approaches one from the positive side okay the limit of f ofx as X approaches 1 from the positive side is just going to be cosine of pi * 1 and thinking back to last year right we can probably take it a little more simply than we did then you know remember Pi is over here cosine is the hor Al displacement from the Y AIS so that looks like negative 1 to me so I'm going to say that's okay negative 1 and then natural log of e * X is the other piece of this piece wise defined function so if I'm thinking X is less than one that will give me information about the limit as X approaches one from the negative side so that liit it's going to equal natural log of e * X and where I guess know I'll just show you me doing that where I plug in x = 1 so be e * 1 and I'm thinking okay natural log of e what exponent do I put on E to get e well yeah I could just do it right there that's e to the first power so that's going to equal one okay so the limit of f ofx as X approaches one in order for it to exist we would have needed to see the same number from the two one-sided limits and we're not seeing the same limit so so I'm going to write out that this limit does not exist and you can write d and e but I need to give a reason okay and really the reason is the disagreement and one-sided limits but that's really not going to be specific enough what you need to write is going to be because of the like you know specifically the one-sided limit from the left is not equal to the one-sided limit from the right and you're going to see in this course you know the whole like explain thing that's going to be just as if not more important than the actual answer itself all right number two is asking us to find the values of the limit as X approaches two from the right and the limit as X approaches two from the left okay so just the two one-sided limits so what we're not going to need is that information about what's actually happening at xals 2 that three and we're not going to need to compare them either to determine if they're equal no if we didn't see a one-sided symbol on there all right so limit as X approaches 2 from the positive side of G of X okay so it's going to be this one X is greater than two so let's say all first this thing is going to equal I'm going to plug in 2 s of 2 piun over 3 and I'm going to draw myself a diagram I remember that 2 pi over 3 it's over there in the second quadrant with a 60° reference angle and if you were in AP pre-cal last year hopefully this is starting to come back this is a thing we did a whole lot um it's a 30 6090 so the legs are one half and ro3 over2 the hypotenuse is one the longer one's roo3 over2 think about what's going to be negative that's going to be this one signs the vertical the ratio of the vertical leg to the you know hypotenuse so it's going to be ro3 over two then the limit as X approaches 2 from the negative side of G of X I'm going to use the other piece okay approaching from the negative side that's where X is less so that's going to equal e 2 - 2 that's e to 0 and we know that that's going to equal one okay okay this next problem this is you know part of why I made a new homework a new set of practice problems I saw this this and I realize that really if I tell you that the limit as X approaches 0 S of 4X overx is equal to 4 uh we can use limit properties to you know explore some algebraic things okay so what is the limit as X approaches 0 of x / s of 4X okay well um what we know to be true is that the limit of one over a thing is going to be one over the limit of that thing just provided that that Li is not going to zero right you know like f ofx is going to two 1 over f ofx is going to go to a half okay and I don't think that's that's too difficult to believe provided that the original limit exists and is not going to zero so in our situation we might be interested in saying okay well this is just the reciprocal of what we saw originally so I mean it's going to go to 1/4 right because this is the limit as X approaches zero of 1 over sin of 4X / X okay and this thing that I'm highlighing is going towards four as X approaches zero stands a reason that this is going to equal 1/4 okay and then in Part B we're going to say what is the limit as X approaches 0 of 3x / tangent of 4X we're going to want to use a trig identity on this and but I don't want you get the wrong idea from this problem we're not going to need to like know a bunch of trig identities the only ones we really need for AP Calculus are the reciprocal identities uh the tangent is s over cosine secant is one over cosine you know ones like that we don't even really need the pag pythagorean identities and if you're an AP preal we really don't need those uh angle Edition formulas that's just not going to be necessary for this course the this one what I'm going to do is kind of break it out okay first we've got this three that isn't there but that's just like kind of a multiplier right so I'm going to say it's 3 over 1 multiplied by okay X over tangent of 4X I'll switch the tangent to S of 4X / cosine of 4X all right so if I'm dividing by that fraction in the denominator the co sign of 4X will kick back to the numerator and know you could take it off to the side and keep some things change some things and flip some things and you see how it works um I'll let you work those details out uh it's going to be the limit as X approaches zero of bring the cosine of 4X over here that'll be good over 1 * X over sin 4X okay and one thing that we know to be true the properties of limits is that the limit of the product is equal to the product of the limits provided that each of the things is is existing individually so this is going to equal the limit as X approaches zero this pen might be too thick but it'll be right 3 cosine of 4X ID by one well I don't need to write that and I know that's going to exist right cuz I can take the cosine of zero and it's not like you know going to be undefined or anything and then I'm going to multiply that by the limit as X approaches Z of X over sin 4X which we actually just computed in part A okay so cine of 4 * 0 okay thinking back to the unit circle z r radians is over here cosin's the horizontal displacement from the Y AIS it's going to be positive 1 so I've got 3 * 1 and then it's the first limit multiplied by okay well I figured out that was 1/4 so that's going to be 3/4s all right now we're going to kind of turn this backwards and we're going to come up with um some drawings of some functions that made a description um I saw this problem set don't remember where but I really liked it um and I think I think it was definitely something that was missing from my last homework so um the limit as X approaches one of this function and the value of the function both exist but the value of the limit is not equal to the value of the function well okay let me draw limit existing in blue so limit existing to me means from the two sides we're approaching the same value okay function itself might not be defined or if it is defined might not be equaling the you know value expect at that whole um and so it's saying a of one exists but the limit is not equal to the value of the function okay well that would just be as long as it was anywhere else um but it was there at xal 1 and so I could just say dra some axes say that's xal one all right neither the limit nor the function exist well I think there's a lot of ways for that to happen um you know you could be creative and TRW that any old way but um couple things come to mind to me a vertical ASM toote is is going to be like okay function is not approaching any one value it's going to infinity or negative infinity and the value of the function itself doesn't exist because youve divide it by zero cause of a vertical ASM til but I think you know thinking about these functions that have these like kind of breaks in them which is kind of keep seeing in the examples I would say this would would be a situation where the limit doesn't exist and just you know provided that I didn't fill in a circle here put in a DOT anywhere you know above xal 2 that would be B of 2 being undefined right there's no actual graph there there's just holes and so that would be that would be a good drawing of that but a vertical ASM toot would do the same or just a blank graph I suppose would hit the mark So C of three exists but the limit doesn't exist hm oh okay well I mean we could kind of start with the same idea the limit doesn't exist so I'm just going to draw you know kind of graph approaching two different things so I've got some here and there um but C of3 does exist so I just need it to be really anywhere um I think it could be here right that would be fine it could be in the at the other spot you know that would just really be the differ and if it was a piecewise function piece-wise defined function the little domain indicators on the right being less than versus less than or equal to or greater than versus greater than or equal to all right uh Part D the limit exists but the function does not okay well it's probably easiest to start these drawing with the limit so the limit is doing one thing you know from both sides we're approaching the same value but the function itself does not exist okay well just as long as I don't fill in that Circle um or fill in anywhere here at xal 4 that that will fit the description all right the limit and the functions value both exist and the limit is equal to the functions value okay that's something that we're going to be talking about uh not next class meeting but the time after that so I'm just going to draw it and now this is something that was I thought was very noticeably absent in AP pre-calculus last year this is a terminology because when I have taught pre-calculus in the past and I mean even college algebra um when I took it uh there was a lot of talk about this word okay but either way the uh functions limit exists and the functions value exists and they're equal okay that's what we would call continuous or what we're going to call continuous here in a couple of lessons um but that might not be something you've heard you know if if I don't know that I've talked about it in like I taught Algebra 2 the year before last and I don't know that we talked about continuous and discontinuous okay number five this does not have anything to do with limits okay I just I stole this problem from somebody else and cuz I really liked it as you know cuz we're right here at the very beginning of our school year so it's never a bad time to practice some of these algebra things these exponent rules because this is actually going to be a pretty big deal in AP Calculus okay so if we're squaring 4X cubed what's going to happen is I'm going to take a 4X cubed and I'm going to multiply it by another 4X cubed okay and that's going to end up being 16 and six total powers of X right because when we're multiplying two things with the same base and different exponents we do add the exponents you can see it on the fours and you can see it on the X okay we're going to see that again four is going to stay the same right there's no other number to multiply the four with 3 powers of X multiply by two more powers of X is going to be a total of five powers of X 1/x we're going to write that as X the ne1 I don't think I necessarily expect you to know what power functions means it's just it's the proper terminology um I just want you to write it you know with exponents instead of fractions and square roots um or at least you know if you're watching this video it would be like oh yeah I remember 1/x^2 that's going to be X the -2 < TK of X is going to be x to the 12 power cube root of x is X the 1/3 power and the cube root of X2 is going to be x^2 taken to the 1/3 power is going to be x^ the 2/3 um you know like like I said doesn't have anything to do with limits like not really at all what we're dealing with in this unit but in the next unit it's going to be a big deal and really after once it becomes a big deal it'll be a big deal for the rest of the course so okay something we need to be well aware of is our exponent rules for AP Calculus all right here we've got a graph of a function calling it f ofx find the limit from the positive side and the value of the limit just in general okay so as X approaches four from above from the positive side see y looks like it is approaching two okay so the limit as X approaches 4 from the positive side side of f is going to equal 2 and then I'm going to say that limit as X approaches 4 just generally if FX one's not existing okay because the two one-sided limits disagree so I'm going to say because fit it in the limit as X approaches four from the negative side is is not equal to the Limit from the right the limit as X approaches 4 from a positive side okay we can see that from the left from the negative side it's approaching four and F of four is equal to four uh but that's not what they ask and they just ask for the value of the limit from the right side and just the limit okay graph of f is shown below find the values of f of 4 well I don't see any graph at xal 4 so I'm going to say f of 4 does not exist exist or is undefined and then the limit as X approaches 4 of f ofx we just need to be thinking okay am I approaching the same thing from the left and the right and I am okay and it looks like x = 4 y = 4 okay so I'm going to say that the limit as X approaches 4 of f of x is equal to four like that four right there I know that might be a little confusing um but you know although F of four is undefined because that hole in the graph uh from both sides the output values f y is approaching four as X gets closer to four okay number eight here we've got a composite limit and this is you know because I am committed to a diversity of problem types and difficulty levels I'm going to show you some ult ones too and this is going to be one that we're going to have to think through okay these composite limits are tricky but I really believe that it's the best thing to do to show these two you right at the beginning let you grapple with them a little bit right now they're not the most important thing I wouldn't even rate them as an important thing at all in August but um I'd let you come back to these problems a couple more times throughout the year and they'll get easier and you'll start to understand it more and you'll see that growth in yourself so that's why we're showing it to you but I don't necessarily need you to be like 100% you know confident on composite limits like this G of G of x f of 1 - x^2 what we've got coming up next um these are tricky I'm just telling you right now so G of G of X as X approaches 2 well let's just think about what's happening on the inside that's what you want to do one step at a time all right so G of x if we just think about what's happening to G as X approaches 2 that is going to equal zero okay and so you would think okay well G of G of X is like kind of getting towards G of zero as X approaches two on the inside of this composite function and that limits not existing so this would be does not exist but if we look more closely G ofx is not just approaching zero it's approaching zero and is exclusively positive it's approaching Zero from above so what's really happening is let me draw the picture so thinking about X approaching two from the left G is going from 1 to a half to a third to a fourth to a really small numberb is getting really close to zero but not actually reaching zero and then from the right as X approaches two from the positive side you know it goes from 3 to 2 and 1/2 to 2.1 to 2.01 um and the output value are getting closer and closer to zero but never actually reaching it so I might say that g of X as X approaches 2 is approaching Zero from the positive side okay it's the best explanation I've got for you for right now I mean I don't want to get too deep into it if you see what I'm saying great if not no big deal so the limit as I'd be happy to discuss it further or not because this is just like you know maybe one question on the quiz maybe one question on the test right okay as X approaches two G of G of X really think about as X is approaching two this thing right here is approaching Zero from above okay so as X approaches zero from above G is approaching four as a result this composite function is going to approach four okay you can make a little table of values and think of G of G of 2.1 g of G of 1.9 and and you know maybe go even a little closer but you would see that the output values um for the interior function would be very close to zero and positive so you know kind of like approaching along this angle and that's that's why we're approaching four okay I'm going to just leave it at that though all right yal F ofx its graph is shown down below what are the values of f of 1 - x^ 2's limit as X approaches zero and F of 1 + x2's limit as X X approaches zero okay U this we just need to think about the inside function because it's algebraic so kind of get you started with a sentence starter here as X approaches zero from the positive side 1 - x^2 is definitely going to approach one right we're getting closer and closer to subtracting zero but if we think about what we're subtracting as X is not equal to zero but close to Zero from the positive side we're going to be subtracting a really small positive number so we're approaching one from Below and then I'm going to say also as X approaches 0 from the negative side 1 - x^2 well I'm going to be taking X as a very small negative number as X approaches Z from the negative side so 1 - x^2 I'm going to take that negative number that's very small and I'm going to square it's going to be even smaller but positive but I'm subtracting from one so this is also approaching one from the negative side so this first limit here here boxing off in blue as X approaches zero of f of 1 - x^2 you know kind of a different way of talking about it say like use a standin variable like call that t 1 - x^2 I'm just going to think of that as T now well that's equivalent to the Limit as T approaches one from the negative side of f of T and that's something we're going to be able to look up from the graph uh that's looking like a half I think you know you don't really know from the graph oh maybe you do yeah you do you do know it's a half because it's on a line segment um for the other one it's going to be very similar except because I'm adding x s and X squ is always a positive number I'm going to always have you know values inside this F right here that are a little bit bigger than one as X approaches zero so really the inside of the function is approaching one one from above so I would say and I'm going to write in the sentence above but just after the fact do it a different way this time the limit as X approaches Z of f of 1 + x^2 is going to be the same as the limit maybe I'll call it U this time the 1 + x^2 as U approaches one from above because 1 + x^2 is always a little bit bigger than one as long as X is not zero f of U and so then I'm going to say that's also something I can look up from the graph and that's going to be as I approach from the positive side the saying I'm approaching negative one and so say right there okay but really the reason for that is I'm thinking about each of the one-sided limits as X approaches zero from the positive side this inside this 1 + x^2 and missed a mark is a approaching one from the positive side and I was saying okay well just let U be the stand in for that that 1 + X2 really it's a function and then also I'm going to point out as X approaches zero from the negative side 1 + x^2 is going to approach one from the positive side as well because I'm still adding the square of a small number which is going to be a positive okay I think I've said enough about this one let's move on okay another composite problem okay I kind of wish I had just stuck to two problems on this homework but you know it's all good we we can keep doing this but if you've seen enough of these and you want to move on to the next page I would totally understand okay the other pages are definitely more important than this one okay so what is the values X approaches Zer of H of H of X well as oh yeah yeah this one this one was particularly tricky I thought when I saw it on the internet um as X approaches zero okay from either side h of X is approaching two okay and so what I would say here is that okay I would think that the limit as X approaches zero of H of H of X that's going to be equivalent to the Limit as X approaches 2 of H of X and that's not existing so that's kind of on my hypothesis but I'm going to need a reason and so I really do need to think about the one-sided limits okay all the tricky problems in AP Calculus with limits they do go back to that definition which is then one-sided limits need to exist and agree all right so I'm going to start by thinking about the right side limit as X approaches zero from the positive side H is approaching two from above so really what's happening to H of H of X is this I'm approaching one okay then um maybe somehow I could indicate that that was later on I don't know first is regular then second is going to be highlighter okay but then the limit as X approaches whoops X approaches zero From Below of H of X okay well as X approaches zero From Below H regular H is approaching two From Below okay so that means I'm now I'm looking at this one that limit's going to equal 4 these are not equal to each other okay and I'm just going to get my answer um this does not exist um I think let me okay yeah okay I can put a not equals there and then I have not said anything that's untrue now the whole not saying anything that's untrue with an equal or I suppose a not equals sign is something that we will discuss more at length in this class because it's kind of like the start of really thinking mathematically in my opinion at least um there are good discussions to have about the the way we present our work but you don't need to worry about that yet I'm just going to say these two things are not equal um but you know this is actually a valid statement okay keep it on going okay we've got a table of values and a request for a sketch of some functions all right so F ofx and it's all near x = 1 so what I'm going to do is I'm just going to draw x = 1 there F of 1 = 4 and the limit of f is 5 okay so that means you know I'm just going to decide that was not successful decide what I want five to be and I'm just going to let F approach that from both sides right because the limit as X approaches one of f exists and equals 5 F of 1 equal 4 that means that the actual value of f is going to be down below okay for G just draw the same thing it doesn't have to be super fancy x = 1 the limit of G is equal to 2 and we kind of already ran into this once with a different type of problem but G of 1 equals 2 so just go right there and H H of one doesn't exist but the limit of H does exist and equals ones and if H1 doesn't exist I don't color in anything so okay and then I we limit property situation okay H ofx multiplied by 2 FX - 3 G ofx all right well all of these limits are existing so we just going to say all right well this is going to be equal to the limit of H ofx which was one that's not a very interesting problem multiplied by 2 * the limit of f which which is 5 - 3 * the limit of G which is 2 K1 - 6 is going to be 4 which was multiplied by one okay so I've got four there right now okay back to some some algebra let's find some values of x to make the equations true these are things that we just need to be able to do in this class and I figured now is a good time to practice and you know maybe give you a break from thinking about those composite limits too hard all right so e 2x equal 5 I want to get that that X that expression variable expression down out of the exponents I'm going to need to use a log and since the base is e use a natural log and I don't know if I said it in your class but you know I've said it many times natural logs really the only log I care about and big part of that is and you know in AP Calculus we really only worry about except in very rare situations uh exponentials with base e and really exclusively Logs with base e so I'm going to take the natural log of both sides a natural log of e 2x well that's what exponent do I put on E to get e to the 2x that's 2X or you could look at it as their you know inverse functions and undo each other but I just like to think about it more fundamentally 2x = the natural log of 5 and then I'm going to divide both sides by two and that's going to be x = 12 of the log of five natural log of five okay and I I don't remember I think that that would maybe that wasn't far enough for AP preal like you had to turn the 1/2 into an exponent of the five and called a square root or something we don't have to do anything with that in this class that one half of the natural log of five that's a perfectly fine answer for us and if we're trying to get rid of a natural log in equation like we see in the middle we're going to use a you know take e to both sides power type trick and it's going to equal e to 4 okay now e to the natural log of x -1 you could think about it you know in terms of what a log means it's the it's a special exponent that that natural log of x - one is the special exponent you put on E to get x - one so if you use that special exponent you should get xus one or you could just say that e and natural log undo each other I feel like that's somehow less less powerful of an understanding so that's 1 + e 4 all right and then this last one is just a nice uh nice second deegree polinomial we know how to solve that we got to start by setting it equal to zero though that's that's my first message for you there's going to be a lot of setting things equal to zero and solving and you know in the event that they are polom we might need to be able to solve them without a calculator but the good news is the types of factoring we have to do really are not that bad in this class it really is just like greatest common factor like the six that's common to all three terms and then you know kind of your oldfashioned Algebra 1 type factoring what two numbers multiply to -2 and add to negative one okay well numbers that multiply to two are two and one so if this is going to work it's going to be T and two I think the two yeah the two would want to be the negative and the one would be positive okay so I've got this factored up I'm mentally checking you know at each step and that's what I recommend you do once you get to the point where you're really confident with your factoring you should factor and then you should mentally like distribute think about and I realize I did it without saying looking at 6t ^2 - 6t - 12 you know I'm going to look at that and make sure I didn't make an error and have to you know embarrass myself on video um so I'm going to just like visually double check and then T minus 2 t + 1 I'm seeing t^2 -2t + 1 t Okay this makes sense minus 2 yeah okay I'm good to go okay so if I'm looking for the solutions okay it's one of these things is equal to Z if their product is zero so either 6 equals 0 which sounds unlikely to me or t - 2 = 0 0 which would happen when T was equal to 2 or maybe t + 1 was equal to 0 and that would happen when T was equal to1 okay so I should this works looking a little crowded so just box the answers for each of the problems and just word about boxing your answer it is not something that you have to do an AP Calculus on the free response it's something that you were able to do it it does impact the scoring but in a way that I would discuss with you if you wanted to to completely separate to what we're doing here all right so the last question we've got is just find the value of the limit as X approaches to for of f ofx for these different functions F and it looks like they're all trig and really I just wasn't going to let you do the first homework of the year without interacting with the unit circle at least a little bit it's not going to be the biggest deal in AP calculus but it's going to be a deal that we have to be able to interact with here and there all right so you know these functions are all nice and smooth have uh good graphs tangent has some vertical ASM tootes but I'm not taking you anywhere near there um I just really want you to figure out some unit circle values all right so the limit as X approaches 2 of F ofx and kind of the thing I the message here is that if there's nothing bad going on with the function near the x value like sine of pi over 4 * 2 that's s of pi over 2 we can figure that out we can just plug in on the function so that's going to be S of 2 piun over 4 which is s of piun two all right well Pi / 2 is up here and that's at the very top of the unit circle which is going to be one unit away from the origin positively away from the X AIS so that's going to equal one okay then for this one um this limit as X approaches 2 of f ofx that's going to be Co cine of 2 piun over3 I'm going to take that first and then I'm going to subtract one okay so I just got to think about where 2 piun over 3 is I think this is actually the second time on this problem set where we've had to interact with 2 pi over 3 um it's up here 60° reference angle so 12 it's going to be negative that's going to be the < tk3 over2 and this will be one okay but I think the last time we were looking for S so cosine is adjacent over hypotenuse -2 / 1 that's -2 minus one and actually this is a safe place to stop for AP calculus but I'm going to keep going cuz I know that this is3 haves 1.5 also works I would really encourage you not to use a mixed number though in this class because if you write negative -1 with a 1/2 next to it that could very easily be read as -1 * a half you think about it so mixed numbers really not not the way to go in this class all right F ofx equal 2 tangent of < / 6X I mean I'm not saying I won't know what you mean but I'm just saying it could be it could be you know very clear and mixed numbers aren't always that so this one's going to be tangent of or two tangent of 2 pi over 6 that's going to be 1 Pi / 3 okay pi over 3 is in the first quadrant that's pretty high up here 60° reference angle okay I've got my diagram it's 30 60 90 so the short leg is a half the long leg is < tk3 over2 and the hypotenuse is 1 everything's going to be positive tangent is opposite over adjacent so < tk3 over2 / a half okay so so two uh time tangent is opposite divided by adjacent so I'll flip and multiply okay cancel off the twos and I'm going to have 2 < tk3 all right and that's going to be all for this problem set I think this is a good representative uh set of problems if you need more work with the limits or or whatever just go look at any of my other videos kind of on this topic um I've got plenty of practice problems okay so that's all for this one thanks for watching