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Limit at infinity of a difference of functions

Sal finds the limit at infinity of √(100+x)-√(x). Created by Sal Khan.
Video transcript
Let's think about the limit of the square root of 100 plus x minus the square root of x as x approaches infinity. And I encourage you to pause this video and try to figure this out on your own. So I'm assuming you've had a go at it. So first, let's just try to think about it before we try to manipulate this algebraically in some way. So what happens as x gets really, really, really, really large, as x approaches infinity? Well, even though this 100 is a reasonably large number, as x gets really large, billion, trillion, trillion trillions, even larger than that, trillion, trillion, trillion, trillions, you can imagine that the 100 under the radical sign starts to matter a lot less. As x approaches really, really large numbers, the square root of 100 plus x is going to be approximately the same thing as the square root of x. So for really, really, large, large x's, we can reason that the square root of 100 plus x is going to be approximately equal to the square root of x. And so in that reality-- and we are going to really, really, really large x's. In fact, there's nothing larger, where you can keep increasing x's, that these two things are going to be roughly equal to each other. So it's reasonable to believe that the limit as x approaches infinity here is going to be 0. You're subtracting this from something that is pretty similar to that. But let's actually do some algebraic manipulation to feel better about that, instead of this kind of hand-wavy argument about the 100 not mattering as much when x gets really, really, really large. And so let me rewrite this expression and see if we can manipulate it in interesting ways. So this is 100 plus x minus x. So one thing that might jump out at you, whenever you see one radical minus another radical like this, is well, maybe we can multiply by its conjugate and somehow get rid of the radicals, or at least transform the expression in some way that might be a little bit more useful when we try to find the limit as x approaches infinity. So let's just-- and obviously, we can't just multiply it by anything arbitrary, in order to not change the value of this expression. We can only multiply it by 1. So let's multiply it by a form of one, but a form of one that helps us, that is essentially made up of its conjugate. So let's multiply this. Let's multiply this times the square root of 100 plus x plus the square root of x over the same thing, square root of 100 plus x plus the square root of x. Now notice, this of course is exactly equal to 1. And the reason why we like to multiply by conjugates is that we can take advantage of differences of squares. So this is going to be equal to-- in our denominator, we're just going to have the square root of 100. Let me write it this way actually, 100 plus x plus the square root of x. And in our numerator, we have the square root of 100 plus x minus the square root of x times this thing, times square root of 100 plus x plus the square root of x. Now right over here, we're essentially multiplying a plus b times a minus b. We'll produce a difference of squares. So this is going to be equal to-- this top part right over here-- is going to be equal to this. Let me do this in a different color. It's going to be equal to this thing squared minus this thing, minus that thing squared. So what's 100 plus x squared? Well, that's just 100 plus x. And then what's square root of x squared? Well, that's just going to be x. So minus x-- and we do see that this is starting to simplify nicely-- all of that over the square root of 100 plus x plus the square root of x. And these x's, x minus x, will just be nothing. And so we are left with 100 over the square root of 100 plus x plus the square root of x. So we could rewrite the original limit as the limit as x approaches infinity. Instead of this, we've just algebraically manipulated it to be this. So the limit as x approaches infinity of 100 over the square root of 100 plus x plus the square root of x. And now it becomes much clearer. We have a fixed numerator. This numerator just stays at 100. But our denominator right over here is just going to keep increasing. It's going to be unbounded. So if you're just increasing this denominator while you keep the numerator fixed, you essentially have a fixed numerator with an ever- increasing, or a super-large or an infinitely-large denominator. So that is going to approach 0, which is consistent with our original intuition.