okay guys let's go ahead and finish up the lesson four notes on finding limits by analytic methods basically means my algebraic methods okay factoring conjugate method that sort of thing we're on down here second page topic 1.9 connecting multiple representations of limits [Applause] okay graphically looking at the function we see that just because it is undefined at a specific x value doesn't mean that we can't find the limit we remember holes okay when we have holes in the graph it's undefined there but we can still find the limit because the graphs all lead back to the same location the hoover dam construction example use the graph of the function to determine the value of each limit below okay so here we have the graph of i assume it's this one they should have told us that this is the graph of they could have named it f of x and right here they just could have put y equals f of x but we're gonna go ahead and say that that's the graph of that even though it should have said it all right so it says the limit as x approaches 2 from both sides of this graph so if i place my pencil on the graph at three let's say and then on the graph at one if i move them both as close as i can to x equals two so it basically seems like i'm there both of those graphs meet at a height of looks like negative i'm going to say negative 2.5 which is what we got up here when we did the cancellation technique negative five over two is negative two and a half okay um four from the right so same graph four from the right so now i need the right side of four which is five i'll take the point at five and i'll follow the graph we have to travel along the ref until we get so close to four that it seems like we're there we can't see it but that's a vertical asymptote right there and this graph's gonna get closer and closer to that vertical asymptote so it's going to be boundless and that's going to go to positive infinity we do label things positive and negative infinity even though they technically don't exist we still put positive or negative infinity you can leave the positive off the infinity and it's still it will still be a positive infinity okay but you definitely have to put the negative on infinity to make sure it's negative infinity but you can leave the positive off if you want to just like you can do with numbers i'm just used to writing them in there all the time okay now four from the left so this time i'll take my pencil and place it on three on the graph right there and then i'll follow the graph until i get so close to four and again you won't be able to see it but we have a vertical asymptote here and this graph is going to get so close to that vertical asymptote never really reaching it though it will never get there and that one goes down to negative infinity the question becomes you've already seen how to figure out the limit at a whole x equals two we can see on this graph there's a hole right there in fact that's exactly what you guys told me when i asked you that extra little question when we did this example what's going on on this graph even though you can't see it you haven't done this problem yet even though you can't see this graph you guys were able to tell me there's a hole at least somebody was able to tell me that there's a hole on this graph at x equals two and now you get to see it but the question becomes how do you find these answers how do you find the answers at vertical asymptotes because if we just try plugging in the 4 it's not going to work again it's going to give us division by 0. so let's see how we can find that we know that there's only two possible answers for a one-sided limit at a vertical asymptote if you have a vertical asymptote there's only two ways it can go they're either going to go up forever or they're going to go down so if i'm looking at this limit which would be the left hand side of let's say this number is c that's that's what she calls it i believe if i want the left-hand limit the left-hand limits either going to be positive infinity or negative infinity same with the other side if i want the right-hand limit at c then that's one of two things for a vertical asymptote it's either going to go up forever give me a positive infinity for an answer or down forever give me negative infinity for an answer okay there are only two possible answers for a one-sided limit at a vertical asymptote let me say that again there are only two possible answers for a one-sided limit at a vertical asymptote either it's positive infinity or it's negative infinity now there are three possible answers for a two-sided limit at a vertical asymptote there are three possible answers for a two-sided limit at a vertical asymptote and let me show you those three possible answers number one possible answer positive infinity from a two-sided limit there's vertical asymptote we'll call that point c again so if your two side of them is going to be positive infinity then the graph of the vertical step is going to go like this up and up from both sides okay so the limit as x approaches c of let's say f of x equals positive infinity because both sides of the vertical asymptote go up but then they can do the opposite both sides can go down at a vertical asymptote you can have the left side go down forever giving you the limit from the left side of negan infinity you can have the right side oh there's that c again go down forever giving you a limit of negative infinity and now the limit as we approach both sides at c uh let's call this g of x since i already named the other one f of x then that's going to equal negative infinity because now you have both sides go down at the very last tip there is one last case one last answer for a two-sided limit at a vertical asymptote the limit as x approaches c from both sides of let's say h of x because it's going to look different than either of those this one is not going to exist and why is it not going to exist because the two one-sided limits will go in completely opposite directions you're gonna have one side go up to positive infinity and it doesn't matter which side but one side will go up and the other side will go down that is also a possibility at a one-sided limit i'm sorry that is also a possibility at a vertical asymptote you can get one-sided limits that go in completely opposite directions one of the easiest rational functions that we have one over x this would be the graph of one over x and this would be the point zero because you can't plug a zero into this function and it gives you a vertical asymptote there okay and i think it would actually go in opposite directions so the one for one over x would look like this because you have negative numbers before zero you have positive numbers after zero this is what the graph of one over x will look like and it's one of the first rational functions that we work with is one over x at least in algebra that's one of the first ones you should have worked with okay so two possible answers for a one-sided limit they're either going to be positive infinity or negative infinity at vertical asymptotes but there are three possible answers for two-sided limits at vertical asymptotes positive infinity negative infinity or they simply don't exist because they go in opposite directions at the vertical asymptote okay now the question is how do we find that without looking at the graph if we're just given the function how are we going to be able to tell well one of the ways is if we were to try plugging in these numbers you guys if we tried plugging in a four and we tried they're both fours doesn't matter which side you approach from so you try put even with one sided limit you guys i'm not sure if i told you this but even with one sided limits the first step you always try substitution so if it asks you to try four from the right always just plug a four in okay now most likely it's not going to work because if it did work then it would it really wouldn't be a calculus question okay but you always still try it so it asks you to try four from the right always just try plugging in a four and i bet you if you plug the four in this bottom is going to give you a zero okay what about the top if i plug the 4 into the top that'd be 16 plus 4 which is 20 20 minus 16 is 14. i didn't get zero on this one you guys i got 14 over zero now i still can't put equals 14 over zero you can't put anything divided by zero as equal to something okay because that is undefined things don't equal undefined okay what about this one same thing on the bottom we try plugging in the four we get zero here we try plugging in the four we'll still get 14 here what about if i plugged in the two into this one this one's the whole remember the hole is right here if i plugged in the two this one would be zero and we already did this one up here and this one would be zero okay i'm not sure if i'm ready to come up with a summary on that for you guys but it's good to know i think okay these were vertical asymptotes at four we have a vertical asymptote at x equals two we have a whole if you try direct substitution into a hole you get zero over zero if you try bird of glass tilt if you try the x value of vertical asymptote you're going to get i'm not sure if you always get a 14. so i'm not sure if i can summarize that for you yet i don't think the number 14 is that important but i do think the number zero is important i'm still a little bit confused i still don't want to give it away about the vertical asymptote yet all right but it looks like a hole you guys is going to give us 0 over 0. all right let's see if we can summarize that in a little bit here let's keep going algebraically finding limits of functions at undefined values consider what happens when you try to evaluate this limit using direct substitution we already just tried that okay which one does it want us to do next page remember when you get indeterminate forms of zero over zero or infinity over infinity your goal remember that's not an answer so many times i've seen people put that down as an answer especially zero over zero your goal is to algebraically manipulate the expression in an effort to remove the point of discontinuity and then try direct substitution again so let's explore a few techniques okay in our example that there's a vertical asymptote at x equals four it's going to be the same function vertical asymptote at x equals four how can we determine the behavior of the function at this x value so this is what we're going to talk about the first thing we're going to talk about is how can we figure out we're not looking at a graph we know we know there's a vertical asymptote there how can we figure out that it's going to be a positive infinity or a negative infinity okay we still don't know really how to tell there's a vertical asymptote there without looking at the graph yet but if we know there's a vertical asymptote there let's figure out how we can tell the answer's gonna be positive infinity or negative infinity okay let's try that so they gave us that there's a vertical asteroid at x equals four we wouldn't know that by the problem yet as x approaches four from the right hand side pick a value to the right of four because when you try plugging this in remember what you get you get 14 over zero right don't put equals 14 over zero because you can't have that that will not make sense this limit is defined 14 over 0 is undefined so the two things should not equal each other they're really particular about what you do with limit statements all right everything else they're really flexible about but limit statements they're really toughy on you okay so let's go ahead and figure out what's going on here let's do this let's factor this since we've already know how to factor or if you guys want some factoring videos i can post those up on your classroom if you guys want them no remember you don't have to be as fast as this but you got to be able to factor these things and then x x we need four and two but they're both going to be negative okay that should be negative two we can't see it too well okay now i can cross out the x minus two with the x minus 2. and the limit as i approach 4 from the right hand side remember we already know that there's a vertical asymptote so we already know that our answer is either a positive infinity or a negative infinity because that's the only possible two answers you can get at it for a one-sided limit add a vertical asymptote so we already know that it's got to be one of two answers our only question now is which one is it is it positive infinity or is it negative infinity to figure this out you guys you don't have to physically do this you don't have to put any number in here okay but i want you to think about a number just to the right of four since we want to the right of four we want a number that's just a little bit bigger than four think of a number a little bigger than four so like 4.1 okay five might be too far away from four so we're not even going to plug this in we're just going to think about this if you think about plugging in a 4.1 in to this function right here all right you've crossed out the x minus 2 you crossed out the x minus 2. if you think about plugging in a 4.1 into this function think about what that is let me get my graph again if i plug in a 4.1 into my function okay that's really close to four now you can even think of my number closer like 4.01 or 4.001 but if you just think about a number that's to the right of four that's really close to it if the graph goes up like this which we know it does if the graph goes up like this we should get a positive number out for that if we plug in a number just to the right of four and if the graph goes up right here we should get a positive number but if the graph went down from the right of 4 like it does to the left of 4 if the graph went down then if we plugged in a 4.1 we should get a negative number okay and that's what's going to help us tell if it's going to be a positive infinity or a negative infinity and the graph is going to go up at the vertical asymptote or the graph's going to go down at the vertebral still but you got to think of a number that's just to the right or just to the left of four if you try a five sometimes you guys sometimes five is too big sometimes going one full number past it is too big okay maybe not for this example in fact definitely not for this example because if i plug in a five i'll get a positive number but sometimes the graph will start below the x-axis and then it goes up really quick at the vertical asymptote so if you were to plug in a 5 sometimes you'd get a negative number and you might think oh that means that the graph's going to go down no we want to think of a number you don't have to plug it in you want to think of a number that's just to the right of four like 4.1 all right and then just think about plugging it in what would you get if you plugged in a 4.1 would you come out with a positive number or a negative number because you already know what the answer is going to be it's going to be some sort of infinity so if i plug a 4.1 into the top and if i add 3 to 4.1 4.1 plus 3 is a positive number if i plug in a 4.1 to the bottom 4.1 minus 4 is still a positive number and then what is a positive divided by a positive a positive divided by a positive is a positive so that tells me that since i already know and remember we haven't figured out how to tell there's a red glass tilt there yet right but if we know there's a vertical asymptote there this is how we figure out if it's going to be a positive infinity or a negative infinity for the one-sided limit and to figure out the two-sided limit add a vertical still you always have to do the one-sided limits first so we have to do this anyway so we know at the limit as x approaches 4 from the right of f x so now she's deciding to name it is equal to positive infinity same thing with this side oh i forgot to put my equal sign right here remember they should always line up even though it's kind of an example and it's a little bit off anyway i should still have an equal sign to say that this limit is equal to this limit or i should copy this whole thing down again right here and just make it a nice long true statement now i'm going to put an equal sign because i'm a little bit too lazy to copy everything now we've already simplified this so we know it's going to equal x plus 3 over x minus 4 because we know we've already factored it and we can go ahead and cancel out the x minus twos okay so now this was from the left side so now think of a number just to the left of four okay like 3.9 or 3.99 so think about uh let's say a number a little smaller than four like three point nine okay not three three might be too far away for us so i'm thinking of it just to the left of 4. so if i plug a 3.9 into the top of my function 3.9 plus 3 is a positive number 3.9 minus 4 is a negative number so now we have a positive divided by a negative we already know our answer is going to be some sort of infinity we don't know which way though in this case a positive divided by a negative is a negative so we know that the limit as x approaches four from the left of f of x which i believe is what she decided to call that is equal to negative infinity so what you just learned is now how to find where the vertical asymptotes are we don't know that yet but what we just learned is how we find the limits at vertical asymptotes without looking at the graph any questions about how we find the limits at vertical asymptotes without looking at the graph yeah i had a question yep so essentially um i get the process but like in terms of like what number to pick when you would plug back in you just pick like any number slightly bigger than whatever it the limit is i do and i always usually just go a tenth 10 to the right a tenth to the left of it you don't go one full unit because in my experience one full unit might not get you the correct sign for your infinity okay all right but one tenth has never let me down okay so that's why i always go 4.1 that's one more tenth and 3.9 is one less tenth and that has never let me down that's great that you understand the process when that's what you first said i kind of didn't get past that because it's the whole process that it can be confusing so that's really good anybody else have a question before we move on this making sense that's really good guys okay now we're going to look at piecewise functions okay this is a piecewise function it's broken up into two pieces piecewise functions can be broken up into many pieces all right piecewise functions are notorious for causing jump discontinuities okay because you follow one part of the graph to one x value and then from that same x value goes a different part of the graph and sometimes these two pieces don't meet at the same height at this this x value all right so i'm not sure how familiar you guys are with piecewise functions okay so let me just give you a little intro to piecewise functions if you ever want to try plugging in a number into a piecewise function let's say you want to plug in the number zero okay you must first check you can't plug them into both pieces because you're not allowed to plug them into both pieces you have to first check they're what i call the gatekeepers these guys right here a lot of people call them the rules that's fine as well there's nothing wrong with that i refer to them as the gatekeepers because they're the ones that allow me to keep going to go through the gate all right so if i want to plug in a zero i have to check through the gatekeeper first so zero would be my mystery number that i'm trying to plug in which is my x my variable so is zero greater than one no it is not so we do not plug the zero into the top piece is zero less than or equal to one it is less than one so if i want to plug a zero into this function i would have to only plug it into the bottom piece so g of zero would equal zero plus one which would equal one so what does that tell me that tells me that the y value on this graph of g when x equals zero is one and that's a little bit about a piecewise function okay they come in different pieces and they're notorious for causing jump discontinuities okay piecewise functions can they're not always i said notorious which means they're famous for it but they don't always do it piecewise functions can cause jump discontinuities and you can start to abbreviate discontinuities if you want d-i-s-c you're going to be using that word quite a bit okay in fact this whole unit's called limits and continuity that's how they tie in okay so find the limits analytically of a piecewise function show your algebraic steps so check this out now this limit wants us to approach one from both sides right we don't have a graph to look at anymore you guys so we have to approach one from numbers that are to the right of one and numbers that are to the left of one just like we would on a graph i go and i put my point at two and i put my point at zero and i move them so close to one that it basically seems like they're at one and i'd say my pencils would meet at the same height that's exactly what i'm gonna do here i'm gonna first think about approaching one from the left-hand side if i think about plugging numbers in that are to the left of one the gatekeepers now are very important all right if i take the limit as x approaches one from the left i can't take one from both sides you guys because there's not one function that will let me take one from both sides there's not one function that says you can use this one for x's that are bigger and less than one no they're split up at one so when i approach one from the left hand side i'm looking at numbers that are just a little bit to the left of one so i'd be looking at zero and then like a half and then like point eight and then point nine and then point nine nine and then point nine nine nine and then point nine nine nine all those numbers are less than one so the function piece that i wanna use is the bottom piece because those are values that i can use when my numbers are less than one so for this function i'm going to put x plus one right there now that i picked the correct piece and that's the whole thing about a piecewise function you don't get to use both pieces at the same time or you don't get to use all the pieces at the same time it depends on what your number you're trying to use is if i have to approach one from the left hand side then that means i have to think about numbers that are just a little bit less than one so i'm using the bottom piece of my function and now that i have the correct piece to figure out this limit even with one sided limit you always try direct substitution first you guys just plug in the one for your missing variable that'll be one plus one and there's no problem with one plus one we've known it since kindergarten one plus one is two it will always be two the weird thing about calculus is i'll go and ask you what two times three is and nobody will want to answer that question anymore because you'll be confused as heck and it'll be way too easy for you i think there's got to be a trick to it huh watch at the end of the year i'll ask you what 2 times 3 is okay now let's try approaching one on this graph from the right-hand side which are numbers that are just a little bit bigger than one so like 2 1.5 1.2 1.1 1.01 1.001 all of those numbers are just a little bit bigger than one so the piece of the function that i use is now the top piece x cubed plus one that's the most important thing about piecewise function is being able to pick the correct piece of the function to use and you always check through the gatekeeper numbers and an understanding of limits was helpful in the fact that we know the left-hand side is approaching from numbers like zero one-half nine-tenths ninety-nine one-hundredths nine hundred ninety-nine one-thousandths and we know numbers that are approaching that are a little bit bigger than one are numbers like two one and a half okay one and one tenth one in one hundredth one and one thousandth and then we pick the correct piece to match that and then we just tried direct substitution plug in the one one cubed plus one one cubed is still one and one plus one is two so now we know at the gatekeeper number you guys now we know that our graphs at one from the left side and the right side both meet at the same height of two which means that we can now find the two-sided limit because we remember the limit existence theorem which says that if the left-hand limit is equal to a number and the right-hand limit is equal to that same number then the two-sided limit is also equal to that number that's a big process with a piecewise function if you're not familiar with the piecewise function you got a little intro to it hopefully you saw those in pre-calculus i know we do a little bit in them in algebra two but we don't do a whole lot in algebra two on piecewise functions okay so my question is since we know that piecewise functions can cause a jump discontinuity my next question to you is is there a jump discontinuity on this graph of g at x equals one you can now tell me from the way the answers look is there a jump discontinuity on this graph at x equals one while you guys are typing in your answer i'm going to graph this for us is there a jump discontinuity on this graph of g at x equals one i will show you how to plug in piecewise functions but for right now that's going to take too long so i'm going to go ahead and just plug it in for us i'm not looking at the screen yet i'm looking at my calculator i'm going to check your answers in a little bit that is good there is no jump or break in our graph there is no jump discontinuity which looks like a break in your graph at x equal to one the two are going to connect even though the two graphs are completely different you guys i'm not sure if you remember what a cube graph looks like but you definitely should know what y equals x plus one looks like this is y equals mx plus b that's a straight line i'm going to graph them for you okay i will teach you how to plug in a piecewise function so that you can graph it but let me show you what this graph looks like here comes the x plus one graph so it gets to one boom and there's the x cubed plus one graph that is a piecewise function because before you get to one on your x-axis you have the line x plus one here is the line x plus one all the way until you get to one once you get to one after one i should say right after one it then goes to a graph of x cubed plus one and i'm gonna show you remember what your x cube graphs look like they look like this and that's what it looks like they don't always meet at the same height you guys the x cubed graph plus one could meet at one spot above it it could be one spot above and there'd be this break in the graph okay for this one though they both made at the same height so there was no break in the graph even though the graphs look kind of weird and these piecewise function graphs always look weird there was no break in the graph there was a limit i could follow it from the left side and from the right side until i get so close to one that they both meet at a height of two okay next piecewise function the limit as x approaches two from both sides on this one we can't use one piece oh wait before i do that you guys i want to talk a little bit about the notation right notice that for my two-sided limit i didn't pick one particular function because there is not one particular function that satisfies a two-sided limit i don't have one function that says oh i can plug in numbers that are a little bit bigger than one and a little bit less than one because that's what a limit does plug the numbers a little bit bigger than one and a little bit less than one but they're so close to one that it seems like they're one all right so there is not one piece that i could pick so what did i use i used the name that it gave me g of x and these piecewise functions will always have a name for you okay so that's why when i put the final limit for my two-sided limit i use g of x because i'm not putting an x plus one and i'm not putting in x plus one i have to use the whole thing same thing with this one when i actually get the final answer for this one i'll use the limit as x approaches 2 from both sides of h of x because it's broken up at 2. i don't have one piece that lets me use just for one piece numbers that are bigger than 2 and less than 2. so i have to approach this from the left side and the right side you can always approach a two-sided limit from the left side and the right side you just don't always have to but you can't always okay now think about numbers that are just to the left of 2 like 1.9 1.99 1.999 all those numbers are less than 2. so we're going to use the top piece remember we check through the gatekeeper we're going to use the top piece of the function x squared minus four x plus seven once we correct once we pick the correct piece of a piecewise function then it's really easy you just plug it in and a lot of times you can just use substitution on these piecewise functions unless they really want to be nasty to us and they put one of the pieces as a rational function and i've seen that too but not yet okay that'll be two squared minus eight plus seven that'll be four plus 7 which is 11 minus 8 is 3. so we know that if we were to graph the left-hand side of this graph we would reach a height of all right let's try the right hand side now we think about numbers just to the right of two like two point one two point zero one two point zero zero one all those numbers are just a little bit bigger than two so now we use the bottom piece of our function for this one and we plug in our two that would be eight i'll just put an eight right there save some room and then minus one remember on this one we do the exponent before you multiply okay so do the exponent think of that as a negative one before you multiply you gotta do the exponents that's two squared that'll be a negative four negative four plus eight is four and four minus one is three so right here we know the left hand side and the right hand side both oh they are they do they both meet at a height of three so just like the other one you guys we could say the limit as x approaches two uh this time h of x is equal to three this one would also not have a jump discontinuity it would not have a break in the graph um well this is for a but if there was a hole at x equals one and the actual point where x was found out was like one there would still be a limit right because it gets to the same place if there's a hole and there's no break in your graph there would still be a limit that is exactly right thank you hey mr herrera yep uh so if there was a jump disc discontinuity in the graph with those but those two numbers like you know how like the first equation equals three and the other one equals three they would be different if there was a jump discontinuity all right even a little bit different if they're a little bit different even there's a jump discontinuity all right thank you no problem good question okay anything else before i move on to the next one okay let's try some more algebra basically what this is okay yep whenever i see rational functions i start thinking factoring okay now factoring or cancellation technique so now we try plugging in the two again whenever i see a rational function like this i always get a little bit nervous about dividing by zero so before i put any equal signs on this see how there's no equal signs on these good job she made that up correctly if i try doing this i wish i plugged it into the bottom first so that's 2 squared which is 4 4 minus 12 is negative eight negative eight plus eight is zero most of the time that's going to happen to us in calculus you guys because if it doesn't then we just plug in the number and we get an answer and that's not really a calculus question you could plug in numbers and null number okay plug it into the top though i want to see what's going on 2 squared is 4 plus 2 is 6 and 6 minus 6 is 0. so we have 0 over 0. i think that's going to be a hole from what i saw let me go back and check the last time we got 0 over 0 let's see here we had a hole and here we have vertical asymptotes we get zero over zero that gave us a hole i wonder if that's always going to work and the vertical asymptotes i just get the zero on the bottom but no zero on top i'm trying to find a pattern i happen to already know it but i'm trying to let you discover a pattern here that lets you decide when you have a vertical asymptote and when you have a hole let's see what happens here okay the answer to this will tell us if there's a vertical asymptote or a hole there so we can always go off the answer as well to tell us but i would kind of like to know before we do it so let's see if we can figure something out here okay so i know that this is going to approach 0 on the bottom and it's going to put 0 on the top so having it equal 0 0 is not going to be correct let's go ahead and factor this and again you don't have to be this good at factoring there's going to be plus three x minus two we've done this one before x minus four and x minus two okay so now we can go ahead and cross out the x minus twos and we end up plugging in the two now that'll be five over negative two we've already done this one negative two and a half and that's good answer though right there now that i see that i have a limit as an answer now i can tell you oh there was a hole there there's a hole in the graph or they call it a point discontinuity i actually like the better name of removable discontinuity and x equals 2. i know that because i got a limit and so far with rational functions there's only two types of discontinuities you can have there's a hole and a vertical asymptote okay the piecewise functions are the ones that give you jump discontinuities and these didn't but they can so rational functions can give you holes or vertical options since we got out a limit since we got a number out to our two-sided limit and it wasn't a positive infinity it wasn't a negative infinity and it definitely existed that means that there's a hole there so here's something i learned if you can't tell if there's a hole or a vertiglass asylum before you do the limit or as you're doing the limit you could definitely tell from the answer you can always tell from the answer to a limit problem whether there's a hole or a vertical asymptote on the graph because i know we're going to get an answer a number when we have a whole we're going to get a positive infinity a negative infinity or a dne and there's a vertical asymptote so this answer of getting a number out tells me there's got to be a hole there i know that division by zero is always going to be a hole or a vertical asymptote a hole or a vertical asymptote okay let's do a new function though here we go again i'm a little bit nervous because i already know what's going to happen when we plug in the 3 which is going to be a 0 at the bottom let's see 9 plus 3 is 12 minus 12 is 0. and on top we get 3 cubed minus 27 that will also be 0. so i'm thinking whole but we can definitely use the answer to find out i'm thinking maybe these zero or zeros are holes and if you're thinking that too let me caution you it's not always the case let's factor it factoring cubes you guys remember how to factor cubes uh that'll be 3 3 3 x minus let's get x plus 3x plus 9. x plus four and x minus three all right again i can get you some factoring videos if you want some help i think it's dan tdm one of them i'm not sure if that's the youtuber that my son watches or is the math guy that i subscribe to uh nancy pai is also another one she does a very good in fact she does a really good algebra stuff um i might put hers up for the algebra for factoring okay um so factored the top factored the bottom i now see that i have a common factor in both the top and the bottom the x minus three so let's go ahead and um cross it out or a wink wink let's go ahead and remove it let's remove the x minus three what do you say well we go ahead and remove it remove it we'll remove the x minus three all right and so now if i try plugging in a three on the top since we removed the x minus three let's try plugging in the three on the top we get three squared plus three times three which would be nine plus nine that's all nines right there nine plus nine plus nine all over three plus four so i'll be 27 over seven and i don't care that simplifies or not because that's my final answer 27 over seven i got an answer out it's not a positive infinity it's not a negative infinity and it definitely existed because i can find it all right and i know that the only two types of discontinuities that we have when we divide by zero are holes which we call point disc connectors or wink wink removable discontinuities removable or the only other option is a vertical asymptote this one's a hole this one is a removable discontinuity i really like the name removable discontinuity when we do algebra this graph will have a hole or a removable discontinuity at x equals 3. we know there's a discontinuity again because of the division by zero if you plug in a number and if you get division by zero that means you're either gonna have a hole there or you're gonna have a vertical asymptote there we know from the answer that this is a hole but maybe you're starting to see a pattern that's why i like the name removable discontinuity okay let's try this one ooh this would be fun to factor okay let's see x approaches two i bet it's gonna go zero zero that'll be two cubed which is eight eight plus four eight sixteen minus twenty six a negative ten plus ten is zero took a little while to figure that out all right two up here eight minus eight is zero plus ten minus ten that'll be zero zero over zero let's see let's try to factor this i'm going to do some side work i don't know if i can factor all that one let me get some paper and i'll factor x cubed so sw is my abbreviation for side work it's work that goes with the problem but you don't want to kind of mess it up i want to factor a four term polynomial now really quick again i'll try to find some factoring videos for you factor by grouping factor out the greatest common factor then factor out the greatest common factor there's always a part that where students get lost so i can factor it to be x minus two times x squared plus five so it's x minus 2 times x squared plus 5. and again when you factor you do not have to be that quick but you got to be able to factor because if you can't factor on this okay next one x cubed plus 2x minus 13x plus 10. yeah that one i don't think i can factor that one not by grouping it's not gonna work like this one does the only thing i know that i can do on this is i'm pretty sure since i got a zero in here you guys this is a heavy level algebra right since i got a zero in here i know there's gotta be a factor of x minus two because if i plug a 2 in here i get a 0. so that means that 2 minus 2 would give me 0 i know there has to be a factor of x minus 2 in there so what i'm going to do is i'm going to divide it out i'll use synthetic division okay i can give you some good news is that on the ap test you probably won't have to do this much algebra definitely won't have to do synthetic division but if you remember synthetic so to factor that one it would be x minus two times it by x squared plus 4x minus 5. now if you're worried about factoring i probably wouldn't worry so much about the factoring on c the factoring that you get mostly on the ap test is a lot faster than that it'll be factorings like you see on a and b but i'm glad you put it in there screw it up yep can we see your work so i can copy it down sure you might have to take a picture of it though because i still need to keep writing and this will be on the video later and again that's just side work so it's kind of sloppy okay thank you no problem again i wouldn't worry so much about factoring c but you definitely should know how to factor things like a and b all right factoring c like i said high level stuff especially the bottom the bottom line is high level okay what i just did on the bottom is high level stuff stuff that you probably won't have to do on the ap test okay they try not to make it so that it's all algebra based they don't want the algebra skills to um they don't want the elder skills to base to be they don't want your grade to be based on the algebra skills they don't want your score to be based on algebra skills they want it to be based on calculus skills so that's why the algebra is not too high level but there's definitely factoring and i would definitely know how to factor a and b if i was you huh c is more high level but now that i have the same factor on top and bottom i can cross that out and i plug in my 2 now i get 2 square which is 4 over two squared plus eight minus five that's going to be nine over that'll be four twelve seven nine sevens so again what does that tell me that tells me that i got a hole or hint wink wink or removable discontinuity at x equals two and the reason i know that is because i removed the x minus two the reason i know i've removed this continuity for this one is because i removed the x minus three the reason i know i've been discontinuity removable discontinuity at two because i removed the x minus two that's why i like the name removable discontinuity because any factor you can remove from top and bottom that's where your hole is going to be so you remove a factor of x minus two on top and bottom then your hole is going to be at two if you remove a factor of x minus three on top and bottom then your whole is going to be at three if you remove a factor of x plus five on top and bottom then your whole is going to be at negative five if you remove a factor of x plus one on top and bottom then your hole is going to be at negative one that's how i can tell whether i have a hole or a vertical asymptote holes are removable discontinuities we can eliminate the problems at holes we can remove the factors that produce those problems x minus two prefer two x minus three for a three x minus 2 for a 2. we can remove those factors that are causing our division by zero okay next up the least common multiple least common multiple method for complex fractions students hate these things okay you probably saw a little bit about complex fractions in algebra two if you try plugging in a zero you get zero here you'll get zero minus 4 which is negative 1 4 and a positive 1 4 will be 0 on the bottom so we get 0 over 0 so we have to be able to work with this somehow let's go ahead and multiply both top and bottom of this by something that will cancel out both of the denominators of my little fractions okay there's different ways to do this you guys so i'm not sure which way you're comfortable doing this but i'll teach you the way that i normally do it i always like to remove the denominators of the little fractions okay i got one big fraction right here but i talk about these as my little fractions and in order to remove those in order to cancel them out i need to be able to have a 4 and an x minus 4. so i'm going to multiply both top and bottom of the big fraction by 4 times x minus four and then when i distribute that on the top i get 4 times x times x minus 4 that's going to be 4 x times x minus 4. that's it if it was 3 times 2 times 7 then that is 3 times 2 times 7. if it's x times 4 times x minus 4 then that is 4 times x times x minus 4 all over now here when i multiply i have to distribute so i'm going to take this and multiply it times the 1 4. so i'm going to be able to cancel out this 4 with this 4 and i'll just have 1 times x minus 4 which is x minus 4 and then when i multiply the 4 times x minus 4 times this little fraction i'd be able to cancel out the x minus fours and i'll just have four okay now i can put the two fours together on the bottom and i get the limit as x approaches zero of four times x over x oh sorry four times x times x minus four over x and if you try using the substitution method right now it's not going to work because you'll still get 0 on the bottom but you can do one thing before you substitute you can cross out the x's they're just being multiplied all right so if i cross out this x up here with this x down here then i have the original limit equal to the limit as x approaches 0 of 4 times x minus 4 and now i can go ahead and use direct substitution on this one because i won't have a division by 0 problem anymore i get 4 times negative 4 which is negative 16. that one's all algebra i'm glad she puts these in there i don't think you'll get one that are that are this complex on your ap test because it's a lot of algebra behind it and usually like to test more of the calculus though but it's good because you need to practice with the algebra i'm sure a lot of you guys are confused about what the heck did you do there yes yeah yeah people hate these fractions see it's not your calculus i'm going to teach you all the calculus you need in order to pass this test it's going to be your algebra skills that hold you back the calculus part is actually the easiest part to teach it's the stuff that you need in order to get to the final answer like the algebra the geometry the trigonometry that's the stuff that can be hard yeah the calculus part is actually the easiest part really all it is is finding the slope of a tangent line all right um okay so this one if i try plugging in the zero i know i get zero here i try plugging the zero here i get one half minus one half so i get zero so again i'm going to go ahead and manipulate this by multiplying both top and bottom by two times 2 plus x the denominators i'm going to multiply both by the denominators of the little fractions i'm going to distribute it to both fractions again to that one and then to this one so when i multiply it times the first fraction the 2 plus x's will cancel out and i'll just have a 2. when i multiply times the second fraction the 2 and the 2 will cancel out and i'll just have a 2 plus x but there's a minus sign in between there so be careful 2 plus x i'll put in parentheses so remember to distribute that minus well it's really good algebra on this stuff all times 2 times x times 2 plus x will be 2 times x times 2 plus x and now we'll distribute the negative sign 2 minus 2 minus x all over 2x times 2 plus x and that equals two minus two is zero so we have negative x over two x times two plus x so we try plugging in the zero for x right now it's not going to work we'll get 0 0 again but we can cross out the x on top with the x on the bottom and now when we try plugging in 0 we'll get negative 1 because that's a 1 up there over 2 times two plus zero that's two so it'll be negative one fourth yeah i told you guys a lot of algebra today but again i'll post this video up online you'll get to rewind it go over this algebra again if you want to get a little bit better understanding of it again it doesn't have to be as fast as i do it but you got to be able to get through some of this okay especially the problems like a and b appear on top you got to be able to factor this this is more complex c and this is definitely complex now could they throw something like this in they definitely throw something like this in but i don't see them throwing something like this in i very rarely see we have to do synthetic division to be able to factor okay next a fun section huh all right so much fun yeah i like it the rationalization technique or kanye you'll love this one i put this on the quiz every year people miss this whenever you the technique of racialization can be used to find the limit when there's a radical in the numerator or denominator okay highlight on that thing right there you only really need this when you have a radical in the numerator or the denominator huh okay the graph of g of x equals the square root of x plus two minus one all over x plus one is shown on the right weird looking graph but that's calculus graph you guys that is definitely a calculus graph the limit as x approaches negative 1 from both sides of that graph well there it is right there i approach negative 1 from both sides that's at a height of i have no idea it looks like about a half but that'd be a guess let's go ahead and find the actual answer this maybe it is a half it looks like it's pretty much a half but let's find the actual answer and let me show you how to do this okay so in order to do this what you want to do if you were to try plug in the negative one i'm just forgetting to do that first because i always know it's going to be a 0 at the bottom at the top it'll also be zero looks like these zero zeros are turned out to be a lot of times holes but not all the time you guys not all the time okay so let's see what can i do to fix this ugly looking thing we're going to do what's called multiplying by the conjugate all right so here's the thing in order to find a conjugate so you can take a conjugate of a binomial anytime you want so if i give you a plus b that's a binomial to find the conjugate of a plus b it's just a minus b that's the conjugate that's how you find the conjugate so if i give you c minus d the conjugate of c minus d is c plus d if i give you x plus five then the conjugate is x minus five you think you got conjugates what if i give you negative x minus four then the conjugate is negative x plus four that's how you find the conjugate all you have to do to find the conjugate is you copy the first term always give me a binomial so you copy the first term don't change the sign on it you just copy it but you change the sign to the second term every single time you change the sign of the second term posit b negative b negative d positive b positive five negative five negative four positive four and that's how you find the conjugate so if i were to give you something like square root of x plus five then the conjugate is square root of x minus five or what if i give you square root of x plus three plus seven then the conjugate is square root of x plus three minus seven think of the radical as one thing that's one thing copy that one thing change the sign to the second what if i give you three minus the square root of x plus four then the conjugate is 3 plus the square root of x plus 4. think of the radical as one thing so you copy the first thing change the sign to the second thing and this is what we're going to need for today because when we do the conjugates it's going to be with radicals so let's go back to the problem we have the square root on top it doesn't matter if it's on top or on the bottom all right we're going to take when we get 0 over 0 like this and we have a square root problem and this doesn't always happen but we get zero over zero like this and we have a square root like this we always multiply by the conjugate of the square root whether it's on the top or whether it's on the bottom we multiply by the conjugate of the square root so the conjugate of the square root of x plus 2 minus 1 is the square root of x plus 2 plus 1. but when you multiply rational function you can't change it so you got to multiply the top and the bottom by the exact same thing you have to multiply by the square root of x plus 2 plus 1. that's the conjugate for the square root of x plus 2 minus 1. and now we're going to multiply the top and the bottom out but here's the good news you don't have to multiply the part that does not have the square root already in it let me repeat that you do not have to multiply out the location of the part that does not have the square root in it okay by location i mean the top or the bottom so in this case the square root was not on the bottom so this is the part that we do not have to multiply out you're just going to copy the bottom down and i would put each little part in its own parentheses so put the x plus one in its own parentheses and then put in the conjugate in its own parentheses but you don't have to multiply them out and that's good news because that would be an ugly foil problem but unfortunately the top you do have to multiply out because the top is going to give you the trick that's going to work for this problem you don't have to multiply the part out that is doesn't have the square root so when we get to problem number b we will not have to multiply out the top this time because the top does not have the square root but you've got to multiply out in this case we have to multiply out the square root part so we have to multiply out the top of this let me do that with some side work so we have the square root of x plus 2 minus 1 times the square root of x plus 2 plus 1. the binomial you guys it's a binomial we're going to foil this out foil f term first square root of x plus 2 times the square root of x plus 2. you might be thinking i have no idea what that is but maybe if i rewrite it for any different way the square root of x plus 2 times the square root of x plus 2 can be rewritten as the square root of x plus 2 quantity squared because there's two of them and now maybe you know what happens when you have a when you square a square root it's going to make it pretty easy for us to work with actually it doesn't look like it's going to be but this is actually going to clean things up so much now here's the part about the conjugate that you're going to like when you multiply the o and the i parts of foil the o and the i parts for conjugates will always cancel each other out that's the trick about multiplying by conjugates you guys the o and the i parts always cancel each other out but to show you that i'm going to multiply them all right the square root of x plus 2 times 1 is positive square root of x plus 2. then we do the o part negative 1 times the square root of x plus 2 is going to be negative square root of x plus 2. so if you have 1 square root of x plus 2's and you're subtracting 1 squared of x plus 2's and those two are just going to cancel out and then you're left with negative one times a positive one which is a negative one so you had the o and the i parts cancel out right here you're going to square a square root you know what happens when you square a square root they cancel each other out they're inverse operations of each other so you're just going to be left with x plus two and then go back all the way over here put that minus one right next to it and then combine those to make x plus one so on the top here you have changed it to be x plus one and that's the whole trick for the conjugate if you multiply by the conjugate it will get rid or cancel out the square roots because you'll always have the square root getting squared so that'll cancel the square root out there and you'll always have the o and the i terms cancel out because that's the whole trick about conjugates the own the i will always cancel out okay now if i look back here i have an x plus one on top i have an x plus one on the bottom let's remove the x plus one from top and bottom and that leaves us with 1 over now by plugging the negative 1 into here i get negative 1 plus 2 which is 1. so we get 1 over the square root of 1 which is 1 plus two is one half just like i thought in the beginning one half but remember we're not going to have the graph to look at anymore we're just gonna have this like our next problem let me do the next problem and then we'll get out of here and i will finish up the rest of the examples on a video and i'll post those up on google classroom as well that way you'll have more examples to look at right here okay but this one i definitely want to go through with you because this one's a hard one like i said i put this up on the quiz every year people miss it every year okay we're gonna multiply both top and bottom by the same thing oh sorry i always tell you i just forget i always just assume it goes to zero over zero because it normally does okay um sometimes the top does not go to zero like we saw fourteen not too long ago so i'm not sure about the top but anytime i get 0 on the bottom i know i got to keep going with algebra so we're going to multiply this by the conjugate so it'll be 3 plus the square root of x plus 4. think of this radical as one thing in this case it's the second thing of our binomial all right the top this time we do not have to multiply out sometimes i also forget to write the limits down every time you're only allowed to not write the limit down after you plug in your number that's when the limit can come off but the limit word lam's got to stay in there until you plug in that number okay so i have to write the limit down again because i'm not ready to plug in the five yet the top i do not have to worry about multiplying i can just copy it exactly like you see it on the bottom this is the one we have to multiply all right let's see if i can do this a little bit faster because we're almost done here 3 times 3 is 9. all right that's the f here comes the o 3 times the positive square root of x plus 4 is 3 square root of x plus 4. here's the i negative square root of x plus 4 times 3 will be negative 3 square root of x plus 4. the o and the i part will cancel out don't worry about them you don't have to write them down if you don't want to huh then you have the l which is a negative times a positive so that's a negative the square root of x plus 4 times the square root of x plus 4 would be the square root of x plus 4 squared so that's just going to be x plus 4. but you got a minus sign right here you guys this is the one that i also put on the quiz i put a minus sign right there because people always forget to distribute the minus sign okay now let's go ahead and keep on working on this again i don't think i'm ready to plug in the 5 yet and if i am i'll just write it down on my next step i'll just copy the top one of the reasons i did not like math at your age because of all the stuff i had to copy when you don't use something in math it doesn't disappear you have to copy it down this would be nine minus x i'm just going to write it right here to save myself a little bit of step and minus 4. so that'll be a negative x minus 4 and 9 minus 4 is going to be 5. that'll be 5 minus x do you guys know what x minus five divided by five minus x is it's not a positive one it's a negative one if i take 5 minus x and if i just rearrange it to look like negative x plus 5 same thing and if i factor out a negative 1 from this i get positive x minus 5. so 5 minus x is really negative 1 times x minus 5. x minus 5 on top x minus 5 on the bottom the x minus 5 will cross out and that just leaves you with negative 1. x minus 5 divided by 5 minus x is negative 1. they're just opposites of each other but the negative one somewhere don't put it on top and the bottom because that'd be a positive one put it somewhere and now i'll try plugging in my five so it'll be three plus the square root of nine over negative one so i'd be three plus three which is six divided by negative one that is negative six and that's the conjugate method you guys not the favorite method of a lot of students it gets a little bit confusing but i would definitely study those two huh i'm going to be putting a quiz up on google forms here pretty soon you guys i'm not sure how much algebra i want to include in it yet but i definitely want to do the graphs finding limits for my graphs okay so if you get your first quiz coming up uh next week okay you guys let me double check the role make sure i got same number yeah we've just lost somebody that's okay i don't blame you that was a long one uh let me go ahead and present no you could drop out anytime you want i told this is the section that most people drop out of you're absolutely right and i never stop anybody from dropping out yep this is a section where we lose a lot of people once the algebra comes in people get scared okay check your roll up you got it if you're there i don't know you guys have a good weekend remember i'll do the rest of those examples i'll put that video up i can stop my recording