Is it possible to determine the limit $$ \lim\limits_{x \to0 } \frac{e^{-\frac{1}{x^2}}}{x^2} $$
I tried using l'Hopital's rule but $\frac{0}{0}$ uncertainty didn't change. Also a degree of function increased. $$ \lim\limits_{x \to0 } \frac{e^{-\frac{1}{x^2}}}{x^4} $$
If I'll use l'Hopital's rule again $\frac{0}{0}$ uncertainty will stay. $$ \lim\limits_{x \to0 } \frac{e^{-\frac{1}{x^2}}}{2x^6} $$ I guess I need other thoughts.
On
Try the arithmetic with a variable switching method :
Let y := $1/x^2$
any devolved to : $$x\;\to\;0\;\;\implies\;y\;\to\;＋\infty$$ because : $x^2 > 0 $
We get : $$\underset{y\rightarrow＋\infty}{\lim} \;ye^{-y}＝\underset{y\rightarrow＋\infty}{\lim} \;\frac {y}{e^y}＝0$$
So : $$\underset{x\rightarrow\;0}{\lim}\;\frac {e^{-\frac {1}{x^2}}}{x^2}＝0$$
If you must use L'Hospital's Rule, here is one way forward. Note that $\frac{e^{-1/x^2}}{x^2}=\frac{1/x^2}{e^{1/x^2}}$ and that $\lim_{x\to 0}e^{1/x^2}=\infty$. Then, we can write
$$\begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}}{x^2}&=\lim_{x\to 0}\frac{1/x^2}{e^{1/x^2}}\\\\ &\overbrace{=}^{LHR}\frac{(-2/x^3)}{(-2/x^3)e^{1/x^2}}\\\\ &=\lim_{x\to 0} \frac{1}{e^{1/x^2}}\\\\ &=0 \end{align}$$
And we are done!