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# Limits at infinity of quotients (Part 1)

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.D (LO)
,
LIM‑2.D.3 (EK)
,
LIM‑2.D.4 (EK)
,
LIM‑2.D.5 (EK)

## Video transcript

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. Now, what I want to think about is-- what is the limit of f of x, 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, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is, as 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-- 4x to the fifth-- is going to become a 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 100 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 4x to the fifth over 6x to the fifth for a 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 canceling out. And so you are left with 2/3. So what you could say 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. 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. So lets 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 2/3. 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 2/3. So let me draw it as neatly as I can. So this right over here is y is equal to 2/3. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2/3. And when we just look at the graph here, 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 negative number, as 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 4x to the fifth and the 6x to the fifth. 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 we're going to get it equaling 2/3. And once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2/3. 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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