Calculating $\lim_{x\rightarrow 0}\frac{1}{x^n}e^{\frac{-1}{x^2}}$ (by hand) $$\lim_{x\rightarrow 0}\frac{1}{x^n}e^{\frac{-1}{x^2}}$$
I tried using de l'Hospital but can't get any further
Thanks!
 A: The key inequality is:
$$e^{y}\geq \frac{y^k}{k!}$$ for any $y\geq0$ and integer $k\geq 0.$ (This inequality follows directly fromn $e^y=\sum_{n=0}^{\infty}\frac{y^k}{k!}$, since all the terms in this sum are non-negative when $y\geq 0.$)
Letting $y=\frac{1}{x^2}$ you can deduce that:
$$0<\frac{e^{-1/x^2}}{x^{2k}}\leq k!$$
Then pick any $k$ so that $2k>n.$ So you get:
$$0<\frac{e^{-1/x^2}}{|x|^n}\leq |x|^{2k-n}k!$$
Then apply the squeeze theorem.

This method proves more generally:

Claim: If $\{a_k\}$ is a sequence of non-negative numbers such that $f(x)=\sum_{k=0}^{\infty} a_kx^k$ converges for all $x\in \mathbb R$ then
  $$\lim_{x\to 0} \frac{1}{f(1/x^2)x^n}=0$$
  if and only for some $k$, $2k>n$ and $a_k>0$.

Proof: 
The "only if" part is easy - if there is no such $k$ then $f$ is a polynomial with $2\deg f\leq n.$ Then $g(x)=f(1/x^2)x^n$ is also a polynomial in $x$, and thus $x^nf(1/x^2)$ converges to a real value as $x\to 0.$
The "if" part is that $f(1/x^2)|x|^n\geq a_k|x|^{n-2k}>0$, so:
$$0<\frac{1}{f(1/x^2)|x|^n}\leq \frac{|x|^{n-2k}}{a_k}$$
and then the squeeze theorem is applied.
When $f$ is not a polynomial (so infinitely many $a_k>0)$ this means that there is always such a $k$ for any $n$ and thus the limit is zero for all $n.$
A: $e^{u}>u^{n}$ for large $u>0$, so for small $|x|>0$, then $e^{-1/x^{2}}\leq x^{2n}$, so $\left|\dfrac{1}{x^{n}}e^{-1/x^{2}}\right|\leq\dfrac{|x|^{2n}}{|x|^{n}}=|x|^{n}\rightarrow 0$. 
A: Let $\frac1{x}=y$ with $y\to \pm\infty$
$$\frac1{x^n}e^{\frac{-1}{x^2}}=\frac{y^n}{e^{y^2}}\to 0$$
indeed $\,\forall n$ eventually $e^{y^2}>y^{2n}$ and
$$0\le \frac{|y|^n}{e^{y^2}}\le \frac{|y|^n}{y^{2n}}\to 0$$
thus for squeeze theorem
$$\lim_{x\rightarrow 0}\frac{1}{x^n}e^{\frac{-1}{x^2}}=0$$
