calculus one video lecture number six on limits at infinity and horizontal asymptotes so we're going to start off by considering the function f of x equals tan inverse of x you should have this graph burned in your brain so we know that this function has two horizontal asymptotes one at pi halves and one at negative pi halves and it also goes through the origin at 0 0 and it's increasing so now we're going to consider limits at infinity meaning the limit as x approaches positive infinity so imagine you're walking on the graph going towards positive infinity tan inverse of x the function approaches what y value well it approaches pi over 2. that's the horizontal asymptote in the positive x direction and similarly if we look at the limit as x approaches negative infinity so imagine you're walking on the graph but this time you're going in the negative x direction now tan inverse of x or the function is approaching negative pi halves and we know that it never actually achieves the values of pi halves or negative pi halves right it's never actually going to equal those values but that's what it's tending towards and those are the equations of the horizontal asymptotes so notice that the equations of the horizontal asymptotes are precisely what the limit of the function is as x approaches positive infinity and negative infinity respectively so this is the definition that we have now for a horizontal asymptote the line y equals l is a horizontal asymptote of y equals f of x if the limit as x approaches positive infinity of f of x equals l or the limit as x approaches negative infinity of f of x equals l now you might remember some method that you learned in precalculus for finding horizontal asymptotes for rational functions dealing with the degree of the numerator and denominator and that's a great way that you can check your work but now that we're in calculus we're going to use this limit definition any time we're asked to find horizontal asymptotes now let's look at this graph here we don't have an equation to work with specifically just yet but i have a function g of x and the graph is shown here in pink so we're going to find the following part a we want to find the limit as x approaches positive infinity of g of x so imagine you're walking on the graph headed in the positive x direction forever and ever what y value or what function value is g of x tending toward well it's oscillating but it does seem to level out and it seems like x is approaching as x approaches positive infinity g of x approaches 2. now what about the other direction what if x approaches negative infinity so now we're walking on the graph we're headed in the negative x direction what y value are we approaching it looks like negative one all right part c the limit of g of x as x approaches two so here's 2 on the graph i need to consider the limit from the left-hand side and the right-hand side separately so let's go ahead and do that so the limit as x approaches 2 from the right of g of x let's see so i'm approaching 2 from the right side from values larger than 2 notice the graph is increasing increasing without bound so that limit is positive infinity and then the limit as x approaches 2 from the left of g of x let's see what's going on so we're coming from the left side of 2 or from values smaller than 2 and now the graph is decreasing without bound the graph or the function values are approaching negative infinity well what does this tell us the limit from the left of 2 is negative infinity whereas the limit from the right of 2 is positive infinity well when the limit from the left does not equal the limit from the right that means the limit does not exist and don't put equals it just does not exist it doesn't equal anything that's what that means lovely now let's look the limit as x approaches zero of g of x so again we're approaching zero let me consider the limit from the left and the limit from the right well in this case i can see that they're both approaching negative infinity the limit from the left equals the limit from the right let's write that out the limit as x approaches 0 from the left of g of x is equal to negative infinity which is also the limit as x approaches 0 from the right so i can say this is equal to negative infinity and then lastly the limit as x approaches negative two from the right so where's negative two negative two is right here we're approaching from the right side so from values larger than negative two this way okay what y value does it seem to be approaching that looks like it's almost negative one-half but not quite so i'll say negative 0.4 just eyeball it looks good last thing we need to give the equations of the asymptotes so that would be horizontal and vertical so let's see what do we have here well i know right away my horizontal asymptotes come from the first two limits that we took the limit to positive infinity and negative infinity so those equations are going to be y equals 2 and y equals negative one and then what about vertical asymptotes did we have any certainly we have one here at x equals zero right remember that limit came out to be negative infinity and we have another one here at x equals positive two very nice so that completes that problem you'll have a few exercises similar to this in your homework where you're just given the graph you're not going to be given the equation that it represents and then you'll have to use it to find some limits and asymptotes now we're going to consider our good friend the reciprocal function f of x equals 1 over x do you remember this guy so let's graph it and then we're going to use the graph to find the following two limits the limit as x approaches positive infinity for 1 over x and again the limit as x approaches negative infinity for 1 over x so reciprocal function has vertical and horizontal asymptotes at the x and y axis i remember learning this in precalc and then just a couple key points that are easy to draw on the graph usually one one i use for this portion here that lives in the first quadrant and then remember it hugs those asymptotes it doesn't cross them and then negative one negative one's another good point to plot for the other portion of the graph beautiful all right now let's use our graph and see if we can evaluate these two limits so the limit as x approaches positive infinity for one over x so now imagine you're walking on the graph you're headed in the positive x direction forever and ever what y value are you tending towards where's the graph going it's going towards zero towards y equals zero and similarly the limit as x approaches negative infinity for one over x let's see now we're walking this way i'm going to put you upside down okay so you don't fall off the graph going this way forever and ever it looks like you're tending towards zero again very good and you should have been able to kind of reason about this limit even without the graph because let's think about what's happening as x approaches infinity the denominator of this expression is getting really really large right so if you think about it x is approaching infinity you have something like one tenth and then it becomes one one hundredth and then one one thousandth and then one over so the larger the denominator becomes the smaller the overall expression becomes so anytime the denominator blows up or it gets large without bound the entire expression approaches 0. and so it doesn't necessarily need to be just 1 over x in fact 1 over x squared 1 over x cubed or one over x to the r for any rational number the limit as x approaches positive or negative infinity will be zero okay so if you have a constant in the numerator and the denominator is blowing up then that means the whole expression is getting really really tiny and so the limit is equal to zero all right so we're going to use that theorem to help us evaluate the following limits and please pay attention make sure when you're working on limits that go to infinity that you show every step of the process just like i'm going to outline here don't skip any steps if you know some shortcut ways and you can mentally check that's good you can use that as a check but if you want full credit on an exam or a quiz or anything you need to show all the steps that i'm going to demonstrate here all right when we're dealing with rational expressions and you're taking limits to infinity you want to divide by the degree of the denominator so the degree of the denominator is determined by the highest power of the variable that you see down there okay sometimes it's x sometimes it's something else so you're going to divide by the highest power of the variable in the denominator in this case the highest power of x that i see in the denominator is to the first so i'm going to divide numerator and denominator by x you'll see why in just a second i want to use the theorem that we just listed so now here i have the limit as x approaches infinity and i'm going to divide all of the terms in the expression by x so 3 divided by x is just or 3x divided by x is 3 and then i have plus 5 over x and then in the denominator x divided by x is 1 minus 4 over x good now this next step this one is a little bit extra and then i want you to understand what's going on you don't have to write it out for every problem i'm going to take the limit now and distribute it to each of the terms in the numerator and denominator so we have the limit as x approaches infinity of three plus the limit as x approaches infinity for five over x and then in the denominator we have the limit as x approaches infinity of minus the limit as x approaches infinity for 4 over x all right let's go term by term so the limit as x approaches infinity of 3 that's just 3. it's a constant it's not going to change fabulous plus limit as x approaches infinity of 5 over x now in the theorem on the previous page we have the limit as x approaches infinity for 1 over x but it doesn't matter if it's a 1 or a 5 or even a million if the numerator is a constant and that constant doesn't change but the denominator is getting bigger and bigger and bigger that's still going to dominate or overpower and the whole expression is going to approach 0. good so we have 3 plus 0 in the numerator now let's look at the denominator so we have the limit as x approaches infinity of 1. i don't see any x's in that expression so that's just gonna be one minus the limit as x approaches infinity of four over x same thing the denominator is getting larger and larger and larger four is just a constant it's not changing so this whole term goes to zero as well and we're pretty much done so in the numerator we have three in the denominator we have one so this limit is equal to three that also is the equation of the horizontal asymptote if we were to try to graph this function now as i mentioned earlier you might remember from pre-calculus that when you're finding equations of horizontal asymptotes if the degree of the numerator is equal to the degree of the denominator all you need to do is take the ratio of the leading coefficients in this case 3 divided by 1 and that gives you the equation of the horizontal asymptote which is 3. but that was cool to do in pre-calculus and now that we are in calculus we justify this procedure or process by taking a limit however use the precalc way to check that you actually did the limit correctly that's fine lovely let's look at another example so here we have the limit as t approaches negative infinity of t squared plus 2 over t cubed plus t squared minus 1. all right remember we need to divide by the degree of the denominator so the highest power of t that i see in the denominator this time is t cubed so i'm going to divide numerator and denominator by t cubed and i have here the limit as t approaches negative infinity t squared divided by t cubed that's going to be 1 over t plus i have 2 over t cubed here and then in the denominator t cubed over t cubed is 1 plus this is going to be 1 over t minus 1 over t cubed now let's go take the limit term by term so as t approaches negative infinity remember if the denominator gets really really large either in the positive or negative direction and the numerator is a constant this whole term is going to approach zero it doesn't matter that it's going to negative infinity 1 divided by negative a million is very small it's just negative instead of being positive like in the previous problem but it's still going to zero all right 2 over t cubed that's also going to zero just faster one is a constant so that's staying one one over t goes to zero and one over t cubed also goes to zero so what are we left with in this limit well in the numerator all i have is zero and then in the denominator i have one and zero divided by one is zero so we're done could we use our pre-calc methods to confirm well do you remember from pre-calculus if the degree of the denominator is higher than the degree of the denumerate the degree of the numerator then automatically the equation of the horizontal asymptote is y equals zero and we confirm that right here our limit was zero which would be the equation of the horizontal asymptote basically the denominator is growing faster than the numerator so the overall limit is going to be zero for the expression good good the only difference notice here with my work compared to the last one is i didn't write lim in front of each term you don't have to do that okay but you do need to show dividing out by the highest power of the variable in the denominator you need to show all of those limits that go to zero and then your result and make sure you write limb as long as you still have variables in that limit okay very good let's consider some more examples that are going to be a little nuanced so you have to be careful here we have the limit as x approaches positive infinity of x plus 2 over the square root of nine x squared plus one now i want you to be careful because do you notice in the denominator how we how we have the square root of nine x squared plus one so what is the degree of the denominator what is the highest power of x that i need to divide by well it's the square root of x squared and that is a highly sensitive little creature that we have to deal with if you'll recall the square root of x squared is equal to the absolute value of x right it's not just equal to x you can't assume that x is not negative and absolute value of x we have defined piecewise as x if x is greater than or equal to zero and negative x or the opposite of x if x is less than zero so we need to be careful here because when i divide out by the highest power of x in the denominator in this case i'm going to divide by the absolute value of x now i have to decide which piece of the piecewise function applies in this limit so what you need to do is look at which way x is going x is going towards positive infinity so i'm headed in the positive x direction which means absolute value of x is defined to be x so i'm going to use the positive portion of my piecewise defined function because this limit x is going to positive infinity i'll show you what you would do if it was going the other way in another example but you do need to show that you thought about it all right now let's go ahead and divide by the absolute value of x and that might be a little bit tricky to do in the denominator so we have the limit as x approaches infinity remember we're just using positive x here in the numerator so i have 1 plus 2 divided by x and then what do we do in that denominator oh my goodness so here we go we'll do this off to the side so i need to divide grab 9x squared plus 1 by the absolute value of x so i need to divide red 9x squared plus 1 by rad x squared okay so you're taking radical 9x squared plus 1 and dividing by x squared so what does this become radical 9 plus 1 over x squared okay so now let's plot that down here in the denominator i have rad 9 plus 1 over x squared and then now we can go ahead and take our limit so as x approaches infinity 1 is 1 2 over x that approaches zero nine is a constant and one over x squared that's going to approach zero so now i'm done taking the limit let's see what's left in the numerator i have just a one and then in the denominator i have square root of nine plus zero which is three so this limit is one third okay these are tricky to work with especially at first so just take your time and make sure you practice enough examples feel free to do extra homework problems do some evens okay here we go example four we have the limit as x approaches negative infinity i have square root of nine x to the sixth minus x over x cubed plus one so you might not be too nervous at first you go okay the degree of the denominator is 3 so i'm going to divide by x cubed but uh oh notice in the numerator again i have radical x to the sixth and well how does that simplify that's the absolute value of x cubed why do i need the absolute value because i don't know if x represents a positive or a negative number i'll tell you when you don't need the absolute value this is probably running through your brain if you have the square root of x to the fourth you don't need absolute value on x squared because x squared's gonna be positive anyways so you don't put them same thing if you had square root of x to the eighth that would come out to be x to the fourth if you raise something to an even exponent you don't need to put absolute value bars on it after right the problem is when you take the square root of x to the sixth or like we did in the previous one when you take the square root of x squared it comes out to be x to the first well if it's raised to an odd power it could be positive or negative and i don't know which way we're going just yet what x represents so you need the absolute value bars around it okay so just check if this exponent is odd then we need them if this was even to start with okay so again we can rewrite this piecewise as positive x cubed if x is greater than or equal to zero or negative x cubed if x is less than zero we'll check which way x is going in the limit to decide which part of the piecewise function are we working with so x is approaching negative infinity that means i'm working with values for x less than 0 in the limit so i'm going to use negative x cubed so how do we deal with that in the limit as follows so since the limit i'm going to summarize right now is going to negative infinity all we need to do is put a negative sign or a minus sign in front of the even root okay so what's that going to look like allow me to show you so we have the limit as x approaches negative infinity where's the even root it's right here you're just going to put that extra negative sign since we're working with the negative portion for how absolute value of x cubed is piecewise defined and then you have radical and then you just move on with your life okay so now we're going to divide by x cubed remember when i put x cubed underneath the radical though it becomes x to the sixth so if you need a little help here you have radical nine x to the sixth minus x and then you're dividing that by x cubed so when you put it underneath the radical now you're going to have nine x to the sixth minus x over x to the sixth okay so in the numerator now we're going to have 9 minus 1 over x to the fifth and then over x cubed divided by x cubed is 1 plus 1 over x cubed okay finally it's time to take the limit so nine is a constant one over x to the fifth that's going to approach zero one is a constant and then one over x cubed that also goes to zero so what are we left with i've got a negative i've got a rad nine and a one those are the only survivors i'll write it out again so we've got negative nine minus zero over one plus zero so this is going to be negative three okay don't fall badly if you're highly uncomfortable from this process it's new most likely so just take your time re-watch some parts that you need to see explained again and just keep practicing okay last example before we change gears we have the limit as x approaches positive infinity look no radicals you can relax a little bit of x cubed minus 2x plus 3 over 5 minus x squared 2x squared so let's divide by the highest power of x in the denominator in this case it's x squared and let's go ahead and see what's going to happen so we've got the limit as x approaches infinity in the numerator x cubed divided by x squared that's just x minus this is going to be 2 over x plus 3 over x squared over we have 5 over x squared minus 2 in the denominator looks good let's go ahead and take that limit now so as x approaches infinity x is going to infinity right okay 2 over x that's going to approach zero three over x squared that approaches zero and five over x squared that approaches zero so what do i have left over after everything goes to zero i've got x over negative two and x is approaching infinity and then if i divide that by a negative two all it's going to do is make the entire expression negative if i take half of infinity or something that's blowing up it's still going to blow up okay it's just going to change in the sense that now it's getting very very large in the negative direction because there's a negative in front of that 2. so for the end result our limit limit's going to be negative infinity good if there was a plus sign in front of the two then it would have just been positive infinity very nice now we're going to look at why sometimes our infinite limits come out to be infinity just like that last problem so we're going to consider f of x equals x cubed another good friend of ours so x cubed let me plot that here some key points 0 0 1 1 2 8 is also a nice one to put 2 4 6 8. so here's 2 8 and then negative 1 negative one and then negative two negative eight that'll be over here and here's x cubed all right now let's use the graph to help us answer this question what is the limit as x approaches positive infinity of x cubed so imagine you're working on the graph headed in the positive x direction forever and ever what y value are you approaching you're not approaching anything the graph is just blowing up or getting larger without bound so that limit is positive infinity now what if you went the other way so the limit as x approaches negative infinity now you're sliding down the graph we so as you go forever and ever in the negative x direction what y value are you approaching the graph is not leveling off it's decreasing without bound so it's approaching negative infinity and sometimes the term that we use to describe this situation is the end behavior you might have heard that from precalculus so anytime you take an infinite limit it tells you about the end behavior of a graph if it has a horizontal asymptote that's the end behavior but other times it doesn't have a horizontal asymptote it either increases or decreases without bound and that information is also revealed by taking an infinite limit or a limit as x approaches infinity okay so let's put this all together so that we have a game plan when we're asked to find asymptotes so if you're asked to find a horizontal asymptote what do you do you're going to take the limit as x approaches positive infinity for f of x and you have to take the limit as x approaches negative infinity for f of x remember not all functions have the same horizontal asymptote in the positive and negative x directions just imagine tan inverse of x that was one that had two different horizontal asymptotes so if when you take these limits if you get a constant right not infinity or negative infinity if you get some constant c then the equation for the horizontal asymptote is going to be y equals c if you're asked to find vertical asymptotes you could take a limit but all we need to do for our purposes is look at the denominator so x equals k is going to be the equation of your vertical asymptote whenever x minus k is a factor of the denominator only meaning it is not x minus k is not a factor of the numerator as well or your expression has to be in lowest terms in order for you to use this and we'll look at an example because that's a very delicate situation okay so first example pretty straightforward find the horizontal asymptotes and vertical asymptotes now we're in calculus so this means we're taking limits so let's start off with the horizontal asymptotes i'm going to take the limit to positive infinity and negative infinity so i've got the limit as x approaches infinity of 1 plus x to the fourth over x squared minus x to the fourth so now i want to divide by the degree of the denominator so i'm going to divide by x to the fourth and let's see what we have here so i have the limit as x approaches positive infinity in the numerator i'm going to have 1 over x to the fourth plus one over this is going to be one over x squared minus one okay so now as x approaches infinity one over x to the fourth is going to go to zero same for one over x squared and the other two terms are just constant so they're going to stay the same and then one divided by negative one gives me negative one all right now would things have changed with my result be different and my limit if x was going towards negative infinity in this case i could see no they would be the same so i'm not going to write down identical work i'm just going to address the fact that i thought about it so we would get the same result as x approaches negative infinity so that means the equation of my horizontal asymptote is y equals negative 1. do our pre-calculus methods confirm this they sure do you would divide the coefficients of the leading terms in the numerator and denominator 1 divided by negative 1. yep that's what we got okay what about vertical asymptotes or asymptotes you could have more than one so we've got y equals one plus x to the fourth over x squared minus x to the fourth now let's factor this completely the numerator doesn't factor but the denominator i can take out an x squared and i'm left with one minus x squared and that actually factors further into a difference of squares so i have 1 plus x to the fourth over x squared times 1 plus x times 1 minus x all right now let's look at the factors of the denominator so we've just got x which is x minus 0 1 plus x and 1 minus x do you see any of those factors repeated in the numerator well no the numerator didn't factor at all so that means the zeros of each of those factors is going are going to be the equations of my vertical asymptotes so my vertical asymptotes are at i have three of them look at us go x equals zero from x squared x equals negative one and x equals positive one so we've got three vertical asymptotes and one horizontal very good all right bonus example are you ready yes so here we go same directions find the equation of the vertical and horizontal asymptotes so we've got y equals x cubed minus six x squared plus eight x over two x squared minus three x minus two all right horizontal asymptote let's start with that so i'm going to take the limit as x approaches infinity of x cubed minus 6x squared plus 8x over 2x squared minus 3x minus 2. so dividing by the degree of the denominator i'm going to divide by x squared and then we'll be able to take the limit so i've got the limit as x approaches infinity in the numerator i'm going to have x cubed divided by x so that's x squared so that's just x minus 6 plus 8 over x and then downstairs what do we got going on we've got a 2 minus 3 over x minus 2 over x squared all right bravo so x goes to infinity that means this term's just blowing up going towards infinity negative 6 is a constant he's going nowhere 8 over x that guy goes to 0 and then 2 that's a constant 3 over x goes to zero two over x squared goes to zero so let's see in the numerator i've got infinity minus six that doesn't matter the minus six eventually as x is blowing up minus six isn't going to do anything divided by two it's still blowing up which way is it blowing up positive or negative direction positive so this limit is going to be positive infinity similarly if we took the limit as x approaches negative infinity let's think what would be different right you would still divide by x squared simplify everything it's just at that last step let's look at the numerator this term would be going towards negative infinity divided by 2 the minus 6 does nothing so that means this limit would be negative infinity okay neither of those are constants though so that means i don't have any horizontal asymptotes in fact notice the degree of the numerator is exactly one higher than the degree of the denominator and that means we have an oblique asymptote so in fact how would we find the equation of that oblique or slant asymptote you would use long division and you could find the equation of the slant or oblique asymptote and this is something you should be familiar with from pre-calculus if you're rusty on long division i'll link up here in the corner a video just going over polynomial long division okay you can't use synthetic for this good so no horizontal asymptotes okay all right that's because there's an oblique now let's look at vertical asymptotes so vertical asymptotes i need to have everything fully factored so let's do that so we've got y equals i notice in the numerator i can take an x out and i'll have x squared minus 6x plus 8 over denominator factors into 2x plus 1 and x minus 2. okay let's keep factoring i have x times x minus 4 times x minus 2 and then in the denominator 2x plus 1 and x minus 2. all right so what does this mean for us in terms of vertical asymptotes because here we have this repeated factor of x minus 2 and we said vertical asymptotes only happen where the factors of the denominator are zero as long as they're not also factors in the numerator so the only unique factor of the denominator is two x plus one so that means my vertical asymptote i've only got one is that x equals negative one-half well you might be wondering what's going on at x equals two because if you look at this graph right the function the domain has two restrictions the domain is x such that x cannot be two or negative one half so what's going on in the graph at two well remember we can cancel out this factor and we actually just have a whole when x is equal to 2. this is a removable discontinuity because we are able to take the limit as x approaches 2. our graph of f of x and we'll have x times x minus 4 over 2x plus 1 the x minus 2 is cancelled out and you get 2 times negative 2. over 5 so negative 4 fifths so you're going to have a hole or an open circle in your graph at 2 comma negative 4 fifths so here's how i think it to myself like both 2 and negative 1 half are not in the domain right we're not allowed to plug in x however since x minus 2 is a repeated factor in the numerator and denominator it cancels out and it's not as dangerous so you just have a little hole in the graph you've got that open circle it's a removable discontinuity but the ones that can't be canceled out after you factor those are super dangerous so those are the ones that are the vertical asymptotes okay and you probably remember doing this a little bit in precalc but without the understanding of limits and continuity and all of what's going on behind the scenes so that concludes the lesson there's a lot of material to digest so just let it marinate take your time and i hope you enjoyed the video please give it a thumbs up subscribe if you haven't already and we've got more limits and graphing to come don't you fret