Finding $\lim_{x\to 0^-} \frac{e^{1/x}}{x}$

Question

I am trying to find the following limit $$ \lim_{x\to 0^-} \frac{e^{1/x}}{x} $$

As $x$ approaches $0$ from the left I see that both the numerator and denominator approach $0$. So this seems like a standard L'Hopital's rule. But when I apply this, then I get $$ \lim_{x\to 0^-} \frac{e^{1/x}}{x} = \lim_{x\to 0^-} \frac{e^{1/x}(-1/x^2)}{1} = \lim_{x\to 0^-} \frac{e^{1/x}}{-x^2} $$ So this doesn't seem to help much. I am a bit lost.

Answer 1

Note that by $x=-\frac1y$ with $y \to +\infty$

$$\lim_{x\to 0^-} \frac{e^{1/x}}{x}=\lim_{y\to +\infty} -ye^{-y}=\lim_{y\to \infty} -\frac{y}{e^{y}}=0$$

indeed eventually $e^y>y^2$ and

$$\frac{y}{e^{y}}<\frac{y}{y^2}=\frac1y\to0$$

Answer 2

Hint:
$\cfrac{\cfrac{1}{x}}{\cfrac{1}{e^{\frac{1}{x}}}}$ Now apply L-Hospital

Answer 3

Let $y=-1/x$, then your limit gives $$-\lim_{y\to\infty}ye^{-y}=-\lim_{y\to\infty}\frac{y}{e^y}=0$$ by l'Hopital's rule