Show that $\lim_{x\to 0}\frac{1}{xe^{1/x}}$ exists

Question

Set $$f(x)=\frac{1}{xe^{1/x}}.$$ I would like to prove $$\lim_{x\to 0}f(x)=0$$ using the definition of limits or results directly related.

What I have tried: Let $(x_n)\subseteq \mathbb{R}$ be a sequence converging to zero with $x_n\neq 0$ for all $n$. Need to show $$f(x_n)\to 0.$$ I am not sure where to go from here. I am not currently asking for a full solution, but a good hint (I would still like to do some of it on my own).

Answer 1

$f(1/n)=\frac{n}{e^n} \to 0$ as $n \to \infty.$

$f(-1/n)=-ne^n \to -\infty$ as $n \to \infty.$

Consequence ?

Answer 2

Hint: Letting $t=\frac{1}{x}$, we have $f=\frac{1}{x}\,e^{-\frac{1}{x}}=te^{-t}$ and we need to study $t\to\pm\infty$. Consider then the inequality $$0<\frac{f}{2}<e^{-\frac{t}{2}}$$ for positive $t$. If $t$ is negative, consider instead $$ f<t\,. $$