Prove that $\lim_{x\to 0} \frac{e^{-\frac{1}{x^2}}}{x} = 0$

Question

Why does $\lim_{x\to 0} \frac{e^{-\frac{1}{x^2}}}{x} = 0$? I tried L`Hospital Rule, epsilon-delta definition of a limit, but none works.

Answer 1

In these ones it is simplest to sub $u=1/x$. Then $x\rightarrow 0^+$ is the same as $u\rightarrow\infty$ and $$\lim_{x\rightarrow0^+}\frac{e^{-1/x^2}}{x} =\lim_{u\rightarrow\infty}ue^{-u^2} =\lim_{u\rightarrow\infty}\frac u{e^{u^2}}=0 $$ by L'Hospital's rule. The limit as $x\rightarrow0^-$ is zero by the same method. It's a common trick when you have a lot of $1/x$'s that you don't want.

Answer 2

$$\frac{1}{x}=u \implies u=\frac{1}{x}\\x \to 0 \space \text{then} \space u \to \infty $$ $$\large \lim_{x \to 0}\frac{e^{\frac{-1}{x^2}}}{x}= \large \lim_{u \to \infty}\frac{e^{-u^2}}{\frac{1}{u}}= \large \lim_{u \to \infty}\frac{u}{e^{+u^2}}\\ \large \lim_{u \to \infty}\frac{u}{1+u^2+\frac{u^4}{2!}+\frac{u^6}{3!}+...} = 0$$

Answer 3

Let $t=x^{-1}$ we get $$\lim_{x\rightarrow 0}\frac{e^{-\frac{1}{x^{2}}}}{x}=\lim_{t\rightarrow \infty }\frac{t}{e^{t^{2}}}\rightarrow 0$$