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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-x\right)$

$\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-x\right)=\frac{1}{2}$

Calculus 1 / AB

Chapter 4

Applications of Derivatives

Section 3

L'Hospital's Rule: Comparing Rates of Growth

Differentiation

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Okay, so here we're taking the limit as X approaches infinity of the square root of X squared, plus eggs minus X. So the real trick about this one is getting it into a form where we can identify that loopy tells rule can even be applied s so we're going to do is take X out as a common factor. So this is equal to the limit as X approaches. Infinity, uh, eggs times the square root one plus one over. Acts minus one. Okay. And I were going to rewrite this a little bit. This is also gonna be equal to the limit. Has X approaches Infinity of the square root of one plus one over X, my ass. 1/1, over X. It was in this form. You consume me, Evaluate it. We put infinity in. We won over infinity on the bottom, and then a one plus one over and face. That should be the square root of one minus one. Which, of course, never run out of room there. But that's gonna end up being equal to 0/0. So that is a correct, indeterminate form that allows us to apply a little beetles rule. So we're going to rewrite our our limit, and it's going to be equivalent to the limit as X approaches infinity. And so we'll start by taking the derivative of the top here. And so I always think of the square roots or any route as being a power to the 1/2. Then we can apply. The power rules of the river is more easily and so the numerator will be equal to 1/2 times one plus one over X to the power of negative 1/2 thanks was attracted one times negative, one over X squared. And so that's inside the derivative from inside the expression here and then the derivative of negative one will become zero. So they have a zero, and then we take the derivative of one over X again, and it's gonna be the same as when we did the inside of the expression, right? So that's going to be negative one over X squared. The nice thing is that we can cancel out this common factor and this can be written as the limit as X approaches. Infinity Ah, 1/2 times one plus one over X to the negative 1/2. So if we evaluate this limit and infinity, we get 1/2 times one plus one over infinity to the negative one house. But this one over infinity is going to be equal to zero. So we really get 1/2 times one plus zero to the negative 1/2. And, of course, one to any power, native or positive is just gonna be one s o. This is equal to 1/2 and that is the solution.

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