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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.
$ \displaystyle \lim_{x\to \infty} \sqrt{xe^{-x/2}} $
0
01:32
Wen Z.
00:34
Amrita B.
Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 4
Indeterminate Forms and l'Hospital's Rule
Derivatives
Differentiation
Volume
Oregon State University
Harvey Mudd College
University of Michigan - Ann Arbor
Idaho State University
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Yeah here we're going to be using we'll have to scroll to find the limit of the function. So we have the square root. Our function is the square root of X. I. M. E. To the um negative X. Over two. So we're just going to rewrite that as the spirit of acts over E. To the Extra Virgin. Um And then that's going to end up giving us since that's X over two, we can just write that as the square root of that. This whole thing is under the square root essentially he to the X. Now that we have this we can plug in infinity as our limit. We see that um that is going to be infinity over infinity. So we take the derivative of the top and bottom and then we get one over infinity. Which is just going to be the square root of zero as a result. And that's gonna give us zero as the final answer.
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