I have to show that $$\lim_{x \to 0}\frac{e^{-1/x^2}}{x}=0$$ by $\epsilon, \delta$
So far, I only have that: Given $\epsilon>0$ such that $|\frac{e^{-1/x^2}}{x}-0|<\epsilon$ then $\frac{e^{-1/x^2}}{|x|}<\epsilon$. But I can't solve for $| x-0 |$ of this expression to obtain a suitable $\delta$
Any one has a hint?
$\textbf{Upgrade}:$
Note that $$|\frac{e^{-1/x^2}}{x}| \leq \frac{\frac{1}{1/x^2}}{|x|}=\frac{x^2}{|x|}=\frac{|x||x|}{|x|}=|x|$$
Then, if $\epsilon >0$ is such that $\epsilon > |x|$. Taking $\delta=\epsilon$ we have that if $|x-0|\leq\delta$ then $|\frac{e^{-1/x^2}}{x}-0| \leq \epsilon$
Is this correct?
$$ \lim_{x\,\to\,0} \frac{e^{-1/x^2}} x = \lim_{u\,\to\,\pm\infty} \frac u {e^{u^2}} \quad \text{(where $u = \dfrac 1 x$)} $$
When $u$ is big (imagine $u=1\,000\,000$ or more?) then when $u$ is incremented by $1,$ then $u^2$ increases by more than $2u$ (more than $2\,000\,000$ in the foregoing example) so $e^{u^2}$ gets multiplied by more than $e^{2u}$ (more than $e^{4\,000\,000}$ in that example). Thus at every such step, the fraction becomes a tiny fraction of what it was before the incrementing of $u.$