Solution of limit $\lim\limits_{x\to 0} \frac {e^{-1/x^2}}{x} $ Small question, I'm trying to solve this limit but I just can't wrap my head around this problem.
$$\lim_{x\to 0} \frac {e^{-1/x^2}}{x} $$
L'Hopital just seems to make it messier. 
It's probably pretty simple - I'd like to hear what I'm missing.
 A: You can write $t=\frac{1}{x}$ so that $t=\frac{1}{x}\to \infty $ as $x\to 0$. Now the limit is $$ \lim_{t\to \infty} e^{-t^{2}}\cdot t=\lim_{t\to \infty} \frac{t}{e^{t^{2}}}=0 $$ 
because $ e^{t^{2}} > t $ as $t\to \infty$
A: The function
$$
f(x)=\begin{cases}
e^{-1/x^2} & \text{if $x\ne0$}\\[6px]
0 & \text{if $x=0$}
\end{cases}
$$
is the classical example of a function which is not the sum of its Taylor series at $0$, because all derivatives at $0$ are $0$. Thus it's not a surprise that l'Hôpital doesn't work in this case.
However, one can do some transformations, for instance trying to compute
$$
\lim_{x\to0}\frac{e^{-1/x^2}}{x^2}
$$
that admits the substitution $t=1/x^2$ which makes it into
$$
\lim_{t\to\infty}\frac{t}{e^t}=0
$$
Thus
$$
\lim_{x\to0}\frac{e^{-1/x^2}}{x}=
\lim_{x\to0}x\frac{e^{-1/x^2}}{x^2}=0
$$
A: From the Maclaurin series we have:
$$x\to 0\,:\;e^x\simeq1+x$$
Subsstitute $x$ by $\frac {-1}{x^2}$ and you will get:
$$\frac{-1}{x^2}\to0\Rightarrow x\to \pm\infty\,:\;e^{\frac {-1}{x^2}}\simeq1-\frac {1}{x^2}$$
So we have:  
$$\lim_{x\to0}\frac{e^{\frac{-1}{x^2}}}{x}=\lim_{x\to\pm\infty}\frac{1-\frac{1}{x^2}}{x}=\lim_{x\to\pm\infty}\frac{x^2-1}{x^3}=0$$
Because the degree of the denominator is more than the degree of the nominator
