Finding a tricky limit.

Question

I am trying find $$\lim_{x \to 0} \frac{1}{\displaystyle xe^{ \frac{1}{2x^2}}}.$$ I think it is zero but I am not sure how to show it.

Answer 1

Set $x^2=1/t$; the limit (for $x\to0$ from the right) becomes $$ \lim_{t\to\infty}\frac{\sqrt{t}}{e^{t/2}} $$ that's really easier.

Just check similarly the limit from the left.

Answer 2

Let $x=1/y$. We want to find the limit of $\frac{y}{ e^{y^2/2}}$ when $y \to \pm\infty$. This limit equals zero.