# `lim_(x->oo) sqrt(x)e^(-x/2)` 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,...

`lim_(x->oo) sqrt(x)e^(-x/2)` 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.

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### Textbook Question

Chapter 4, 4.4 - Problem 42 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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As `x->+oo,`

`sqrt(x) -> +oo` and `e^(-x/2) -> 0.`

So the initial expression is an indeterminate form `oo*0.` Let's transform it to use l’Hospital’s Rule:

`sqrt(x)*e^(-x/2) = x^(1/2)/((e^x)^(1/2)) = (x/e^x)^(1/2).`

Consider the base, `x/e^x.` It is infinity over infinity, everything is differentiable and `(x')/((e^x)') = 1/e^x -> 0` as `x->+oo.`

So the l’Hospital’s Rule is applicable and limit of the base is 0. Therefore the base at the power 1/2 also has limit of 0. This is the answer.

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