Evaluating $\lim\limits_{x\rightarrow0}\frac{e^{-1/x^2}}{x}$ Today in my analysis class, we were preparing for the final and this question came up:

Evaluate $$\lim_{x\to0}\frac{e^{-1/x^2}}{x}$$

We tried taking the $\log$, using L'Hopitals and some other tricks but couldn't figure it out. 
I thought maybe viewing this limits as a sequence in the following way might help;
$$\lim_{x\rightarrow0}\frac{e^{-1/x^2}}{x} = \lim_{n\rightarrow \infty}\frac{e^{-n^2}}{1/n} = \lim_{n\rightarrow \infty}\frac{n}{e^{n^2}}$$
But from them I'm not sure. 
Thank you.
 A: We can write
$$
\lim_{x \to 0} \frac{e^{-1/x^2}}{x} = \lim_{x \to \infty} \frac{x}{e^{x^2}}
$$
and then use de l'Hospital to obtain
$$
\lim_{x \to 0} \frac{e^{-1/x^2}}{x} = \lim_{x \to \infty} \frac{1}{2xe^{x^2}} = 0.
$$
A: Git Gud’s comment (now deleted) is right, but you have to be a little careful when you change the variable. You have 
$$\lim_{x\rightarrow0}\frac{e^{-1/x^2}}{x}=\lim_{x\to 0}\frac1{xe^{1/x^2}}\;,$$
and you’d like to substitute $u=\frac1x$ and take the limit as $u\to\infty$. However, $u\to\infty$ only if $x\to 0^+$; if $x\to 0^-$, then $u\to-\infty$, so I do two calculations:
$$\lim_{x\to 0^+}\frac1{xe^{1/x^2}}=\lim_{u\to\infty}\frac{u}{e^{u^2}}\overset{*}=\lim_{u\to\infty}\frac1{2ue^{u^2}}=0\;,$$
and
$$\lim_{x\to 0^-}\frac1{xe^{1/x^2}}=\lim_{-u\to\infty}\frac{-u}{e^{u^2}}\overset{*}=\lim_{u\to\infty}\frac{-1}{2ue^{u^2}}=0\;.$$
(In each calculation the starred equality is from l’Hospital’s rule, though you should recognize without it that $-$ as Git Gud said $-$ the exponential in the denominator crushes the numerator.) Since the two one-sided limits are equal, the original limit is also $0$.
A: As you noted, it sufficed to compute $\displaystyle \lim_{n\to\infty} \frac{n}{e^{n^{2}}}$. Well, using L'Hopital and the squeeze theorem and whatnot,
$$
0 \leq \lim_{n\to\infty} \frac{n}{e^{n^{2}}} \leq \lim_{n\to\infty} \frac{n}{e^{n}} = \lim \frac{1}{e^{n}} = 0.
$$
