# Evaluate a limit

I've been stuck on this limit $$\lim_{x\to 0} \frac{e^{-1/x²}}{x²}$$ i've tried evaluating from both sides, and got $$\lim_{x\to 0^-}\frac{1}{e^{1/x²}}\times \frac{1}{x²}$$ which actually gives 0 times something that tends to infinity. Same thing happens with the right hand limit, not sure from where to go from here, i don't see any possible cancellations or anything to get rid of 1/x².

• Let $u=1/x^2$. Then this limit is $\lim_{u\to\infty} ue^{-u}$. Could you finish from here? Nov 15, 2020 at 0:44
• Doesn't that give (something that tends to infinity)*(0)? Which is undefined? Nov 15, 2020 at 0:47
• $\infty\cdot0$ is an indeterminant form, yes. Just like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Have you learned about how to evaluate any limits that involve indeterminant forms? Nov 15, 2020 at 0:49
• I mean yes, but i'm not sure on this one, it has nothing to factor/no special limits, not sure. And i prefer not to do l'hopital Nov 15, 2020 at 0:52
• You used the tag limits-without-lhopital. I would recommend looking up L'Hopital's rule because I am not sure how to do this without it. Unless you happen to know about Taylor Series. Nov 15, 2020 at 0:57

$$L=\lim_{x\to 0} \frac{e^{-1/x^2}}{x^2}$$ Let $$x=1/t$$, then $$L=\lim_{t\to \infty} t^2e^{-t^2}=\lim_{t \to \infty} \frac{t^2}{e^{t^2}}\to \frac{0}{0}$$ Use L-Hospital, then $$L=\lim_{t \to \infty} \frac{2t}{2t e^{t^2}}=\lim_{t\to \infty} e^{-t^2}=0$$
$$e^x=\underset{n\to \infty}\lim\left(1+\frac xn\right)^n$$. It's easy to show that the sequence is increasing so $$e^x\ge \left(1+\frac xn\right)^n\ge 1+x.$$ So, $$1+x\le e^x$$ and so $$0\le e^{-x}\le \frac{1}{1+x}.$$ Thus, $$0\le e^{-x^2}\le \frac{1}{1+x^2}\Rightarrow 0\le x^2e^{-x^2}\le \frac{x^2}{1+x^2}$$ which implies that $$x^2e^{-x^2}\to 0$$ as $$x\to 0.$$ But this is exactly the limit you want, once you make the substitution $$t=1/x.$$