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?

Actually, this limit can be done through a simple substitution. As indicated we assume $x\to0^+$ for your limit. Setting $x=1/t$, the limit becomes $\lim \limits_{t \rightarrow \infty} \frac{t^n}{e^t}$. After applying L'Hospital's Rule sufficient amount of times (it's an infinity over infinity situation), your polynomial numerator gets "exhausted" whereas the e-power in the denom stays "as is". And thus the limit goes to zero!  
