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I'm having a hard time figuring out this limit problem: $$\lim\limits_{x \to \infty}{e^{-x}\sqrt{x}}$$

I know that as $x \to\infty$, $e^{-x}=0$ and $\sqrt{x}=\infty$.

My reasoning from here is that since $0(\infty)$ is an indeterminate form, I can rearrange $e^{-x}\sqrt{x}$ as $$\frac{e^{-x}}{\frac{1}{1\sqrt{x}}}$$ and apply L'Hopital's rule. But after I try it, I end up with something that's still indeterminate (this time of form $\frac{0}{0}$), and repeating the process doesn't seem to be helping. Thoughts/explanations?

Asker
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  • Hint: Using the power series for $e^x$. we conclude that if $x$ is positive then $e^x\gt 1+x\gt x$. We don't really need the power series, for let $f(x)=e^x-(1+x)$. Then $f(0)=0$ and looking at $f'(x)$ tells us that $f(x)$ is increasing. – André Nicolas Nov 25 '14 at 00:33

4 Answers4

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Do it the other way and apply L'Hospital rule, i.e., write the expression as $\dfrac{\sqrt{x}}{e^x}$.

Adhvaitha
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Also you can use that $e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$.

So $0< \lim_{x\to\infty}\frac{\sqrt{x}}{e^x} <\lim_{x\to\infty}\frac{\sqrt{x}}{1+x}=0 $

So its limit is $0$.

Martin Sleziak
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Paul
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$$ \lim\limits_{x \to \infty}{e^{-x}\sqrt{x}} = \lim\limits_{x^2 \to \infty}{xe^{-x^2}} $$ but $$ \int_0^{\infty}xe^{-x^2} dx = \frac12 $$ the non-negative $C^{\infty}$function $xe^{-x^2}$ is decreasing for $x \gt \frac12$.

if $\exists \epsilon \gt 0$ such that $ \forall x \gt \frac12$, we have $ xe^{-x^2} \gt \epsilon$ then the integral would not converge

David Holden
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Using L'Hospital, $0 = \lim_ {x \rightarrow \infty} \Big (\frac {1}{2e^{2x}}\Big)= \lim_{x \rightarrow \infty}\Big(\frac {x}{e^{2x}}\Big)= \lim_ {x \rightarrow \infty} \Big(\Big (\frac {\sqrt x}{e^{x}}\Big)^2 \Big).$ Hence, $\lim_ {x \rightarrow \infty} \Big (\frac {\sqrt x}{e^{x}}\Big) = \sqrt{0}=0.$

some person
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