limit ( e^(-1/x^2)/x,x,tends to infinity)
Here's a method. I hope it will suffice. Kind of one sided though.
$\displaystyle \lim_{x\rightarrow{0}}\frac{e^{\frac{-1}{x^{2}}}}{x}$
Let $\displaystyle t=e^{\frac{-1}{x^{2}}}$
Then you get:
$\displaystyle \lim_{t\rightarrow{0^{+}}}t\sqrt{-ln(t)}=0$
You can always try the "ol' Hospital" rule.
L'Hôpital's rule seems to be a dead end for this problem. The second form ends up as $\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{( - 1)( - 2)x^{ - 3} e^{\frac{{ - 1}}{{x^2 }}} }}{1} \implies \mathop {\lim }\limits_{x \to 0} \frac{{2e^{\frac{{ - 1}}{{x^2 }}} }}{{x^3 }}$. I don't know if there is any analytical way to solve this without the use of a substitution.