How can I show that $\lim_{x \to 0}\frac{e^{-\frac{1}{x^2}}}{x}=0$ by $\varepsilon, \delta$
Solution 1:
$$ \lim_{x\,\to\,0} \frac{e^{-1/x^2}} x = \lim_{u\,\to\,\pm\infty} \frac u {e^{u^2}} \quad \text{(where $u = \dfrac 1 x$)} $$
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 $u.$