What is $\lim_{x\to 0} (1+x)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)}$

Question

$\lim_{x\to 0} (1+x)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)}$

Obviously L'hopital is inapplicable here. I guess it can be done by saying that $e^{-\infty}$ is almost zero so the limit is zero but what's the formal way of doing this?

Answer 1

Note that $$ \frac{1}{|x|} + \frac{1}{x} = \begin{cases} \frac{2}{x} & \mathrm{if\;} x > 0 \\ 0 & \mathrm{if\;} x < 0 \end{cases} $$ In consequence, one has $$ \lim_{x\to0^+} (1+x)e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)} = \lim_{x\to0^+} (1+x)e^{-2/x} = 0 $$ but $$ \lim_{x\to0^-} (1+x)e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)} = \lim_{x\to0^-} (1+x) = 1, $$ so that the limit doesn't exist in the end.