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Current time:0:00Total duration:4:07

AP.CALC:

LIM‑2 (EU)

, LIM‑2.D (LO)

, LIM‑2.D.3 (EK)

, LIM‑2.D.4 (EK)

, LIM‑2.D.5 (EK)

so we have f of X equaling 4x to the fifth minus 3x squared plus 3 all of that over 6x to the fifth minus 100x squared minus 10 and what I want to think about is what is the limit of f of X as X approaches as X approaches infinity and there are several ways that you could do this you could actually try to plug in larger and larger numbers for X and see if it seems to be approaching some value or you could reason through this and when I talk about reasoning through this is to think about the behavior of this numerator and denominator as X gets very very very large and what I'm talking about that what I'm saying is is X gets very very large let's just focus on the numerator as X gets very very large this term right over here in the numerator for X to the fifth is going to become much much more significant than any of these other things something squaring gets large but something being raised to the fifth power gets raised that much much faster similarly in the denominator this term right over here the highest degree term 6x to the fifth is going to grow much much much faster than any of these other terms even though this has a hundred as a coefficient or a negative 100 as a coefficient when you take something to the fifth power it's going to grow so much faster than x squared so as X gets very very very large this thing is going to approximate for X to the fifth over over 6x to the fifth over 6x to the fifth for very large for very large large X or we could say as X approaches infinity now what could this be simplified to well you have X to the fifth divided by X to the fifth these are going to grow together so these you can think of them as cancelling out and so you are left with two thirds so what you could say is is the limit of f of X as X approaches infinity as X gets larger and larger and larger all of these other terms aren't going to matter that much and so it's going to approach 2/3 it's going to approach 2/3 now let's look at the graph and see if that actually makes sense what we're actually saying is that we have a horizontal asymptote at y is equal to 2/3 look at the graph so right here is the graph got it from Wolfram Alpha and we see indeed as X gets larger and larger and larger f of X seems to be approaching this value that looks right at around two-thirds so it looks like we have a horizontal asymptote right over here let me draw that a little bit neater we have a horizontal asymptote right at two-thirds so let me draw it as neatly as I can so this right over here is y is equal to Y is equal to two-thirds the limit as X gets really really large as it approaches infinity y is getting closer and closer and closer to two-thirds now when we just look at the graph here it seems like we haven't it seems like the same thing is happening from the bottom direction when X approaches negative infinity so we could say the limit of f of X as X approaches negative infinity that also looks like it's 2/3 and we can use the exact same logic when X becomes a very very very very negative number as it gets if it becomes further and further to the left on the number line the only terms that are going to matter are going to be the 4 X to the 5th and the 6th X to the 5th so this is true for very large X's it's also true for very negative X's so we could also say as X approaches negative infinity this is also true and then the X to the fifth over the X to the fifth is going to cancel out these are the dominant terms and you were going to be get and we're going to get it equaling two-thirds and once again you see that you see that in the graph here we have a horizontal asymptote at y is equal to two-thirds whether we take the limit of f of X as X approaches infinity we get 2/3 and the limit of f of X as X approaches negative infinity is 2/3 so in general whenever you do this you just have to think about what terms are going to dominate the rest and focus on those

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