How to prove that $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x} = 0$ Today someone asked me how to calculate $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x}$. At first sight that limit is $0$, because the exponential decreases faster than the lineal term in the denominator. However, I didn't know how to prove it formally.
I thought of expressing the exponential with its series expansion, but the person who asked me this didn't know about series expansions yet. Niether does he know about L'Hopital.
So, is there a "simple" way to prove that $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x}=0$?

Putting $t = 1/x$ we see that $t \to \infty$ as $x \to 0^{+}$. Also the function is transformed into $t/e^{t}$. Next we put $e^{t} = y$ so that $y \to \infty$ as $t \to \infty$. Thus the function is transformed into $(\log y)/y$. Since $y \to \infty$ we can assume $y > 1$ so that $\sqrt{y} > 1$. We have the inequality $$\log x \leq x - 1$$ for all $x \geq 1$. Replacing $x$ by $\sqrt{y}$ we get $$0 \leq \log\sqrt{y} \leq \sqrt{y} - 1 < \sqrt{y}$$ or $$0 \leq \log y < 2\sqrt{y}$$ or $$0 \leq \frac{\log y}{y} < \frac{2}{\sqrt{y}}$$ for $y > 1$. Applying squeeze theorem when $y \to \infty$ we get $$\lim_{y \to \infty}\frac{\log y}{y} = 0$$ and therefore the desired limit is $0$.
