Proof that $\lim_{x\downarrow 0}x^me^{\frac{-1}{x}} =0,m\in\mathbb{Z}$ with L'Hospital For the case that $m\geq0$ I don't need to apply L'Hospital.
Let $m<0$
We have $x^m=\frac{1}{x^{-m}}$
We also know that $x^{-m}\rightarrow 0$ as $x\rightarrow 0$
We also know that $e^{-\frac{1}{x}}<\epsilon\iff x<-\frac{1}{\ln \epsilon}$ 
Therefore:  $e^{-\frac{1}{x}}\rightarrow 0 $ as $x\rightarrow 0$
Since $x^m$ and  $e^{-\frac{1}{x}}$ are both smoth (infinitely times differentiable) in $\mathbb{R^+}$ I can use L'Hôspital.
I have got the hunch that I have to use L'Hospital $-m$ times. But I don't know how the Expression would look like then.
Here is what I have tried to calculate the first derivative:
$(\frac{e^{-\frac{1}{x}}}{{x^{-m}}})^{'}=\frac{e^{-\frac{1}{x}}\frac{1}{x^2}x^{-m}-e^{-\frac{1}{x}}(-m)x^{-m-1}}{{x^{-2m}}}=x^{-m}\frac{e^{-\frac{1}{x}}\frac{1}{x^2}-e^{-\frac{1}{x}}(-m)x^{-1}}{{x^{-2m}}}=\frac{e^{-\frac{1}{x}}\frac{1}{x}-e^{-\frac{1}{x}}(-m)}{{x^{-m-1}}}=e^{-\frac{1}{x}}\frac{\frac{1}{x}+m}{x^{-m-1}}=e^{-\frac{1}{x}}\frac{1+xm}{x^{-m}}=\frac{e^{-\frac{1}{x}}}{{x^{-m}}}+\frac{me^{-\frac{1}{x}}}{{x^{-m-1}}}$
But this gets me nowhere because I did not get rid off a power of $x^{-m}$
Please help me to figure out where the Problem is and what the term would ook like after I have differentiated it $m$-times.

With $m:=-k<0,\,y:=1/x$ we want $\lim_{y\to\infty}y^k\exp -y$. This is $0$ because $\int_0^\infty y^k\exp -y dy=k!$ is finite.
