Using L'Hopital's Rule to show limit is 0? I am trying to show for any non-negative integer $n$, $\lim \limits_{x \rightarrow 0+} \frac{e^{-{1 /x}}}{x^n}=0$.  For $n=0$ this follows directly since $1/x \rightarrow +\infty$ as $x\rightarrow 0+$.  For $n>0$, I notice the limit has indeterminate form $0/0$ but applying L'Hopital's Rule directly gives $ \lim \limits_{x \rightarrow 0+} \frac{e^{-{1 /x}}}{nx^{n+1}}$, which only seems to complicate the problem.  Also, it seems that repeatedly applying the rule would just increase the exponent of $x$ on the denominator.  Why doesn't L'Hopitals Rule not work directly in this problem, and how could it be used, if at all, to evaluate the limit?  Alternatively, is there another method to evaluate it?

No L'Hospital's rule required here: setting $u=\dfrac1x$, it comes down to showing $\;\displaystyle\lim_{u\to +\infty}\dfrac{u^n}{\mathrm e^u}=0 $, which is a result from high school.
