How to compute $\lim_{x \to \infty} \sqrt{x}(e^{-1/x}-1)$?

Question

I need to compute $$\lim_{x \to \infty} \sqrt{x}(e^{-1/x}-1)$$ using L'Hôpital's rule. Please hint me how to do it.

Thanks!

Answer 1

Let $\dfrac1x=h$

$$\lim_{h\to0^+}\frac{e^{-h}-1}{\sqrt h}$$ $$=\lim_{h\to0^+}\frac{e^h-1}h\cdot\frac{\lim_{h\to0^+}\sqrt h}{-\lim_{h\to0^+}e^h}=\cdots$$

Answer 2

$$\lim_{x \to \infty} \sqrt{x}(e^{-\dfrac{1}{x}}-1)=\lim_{x \to \infty}\frac{(e^{-\dfrac{1}{x}}-1)}{\frac1{\sqrt{x}}}$$

Answer 3

You actually don't need l'Hôpital's theorem: if you use $t=1/x$, the limit becomes $$ \lim_{t\to0^+}\frac{e^{-t}-1}{\sqrt{t}}= \lim_{t\to0^+}\frac{e^{-t}-1}{t}\sqrt{t} $$ and the limit of the fraction is the derivative of $x\mapsto e^{-x}$ at zero, which is $-e^{0}=-1$. So the limit is $0$.

With l'Hôpital's rule without substitutions, you need to do some algebra: $$ \lim_{x \to \infty} \sqrt{x}(e^{-1/x}-1)= \lim_{x \to \infty} \frac{e^{-x^{-1}}-1}{x^{-1/2}} \overset{(\mathrm{H})}{=} \lim_{x \to \infty} \frac{e^{-x^{-1}}(-x^{-2})}{(-1/2)x^{-3/2}} $$ Simplify and you're done.