$\lim\limits_{x\to0}{\frac{e^{-1/x^2}}{x^2}}$ without L'Hopital

Question

How do I compute this limit $$\lim_{x\to0}{\frac{e^{-1/x^2}}{x^2}}$$ without using L'Hopital.

Answer 1

If you set $1/x^2=t$, the limit becomes $$ \lim_{t\to\infty}te^{-t}=\lim_{t\to\infty}\frac{t}{e^t} $$ Now $e^t=1+t+t^2/2+\dots>1+t+t^2/2$.

Answer 2

Another approach is to prove by induction on the degree that $$ \lim_{x\to 0}P(1/x)e^{-1/x^2} = 0 $$ for all polynomials $P$ with $P(0)=0$.