Proof that $\lim_{x\downarrow 0}x^me^{\frac{-1}{x}} =0,m\in\mathbb{Z}$ with L'Hospital

Question

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.

Answer 1

Direct application of L'Hospital's Rule does not provide a tractable way forward as mentioned in the OP.

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To see this, we begin by writing (for $m<0$, $|m|\in\mathbb{N}$)

$$\begin{align} \lim_{x\to0^+}\left(x^me^{-1/x}\right)=\lim_{x\to0^+}\left(\frac{e^{-1/x}}{x^{|m|}}\right)\tag1 \end{align}$$

But, differentiating $|m|$ times, we find that

$$\begin{align} \lim_{x\to0^+}\left(\frac{e^{-1/x}}{x^{|m|}}\right)&=\lim_{x\to0^+}\frac{P_m(1/x)e^{-1/x}}{(|m|!)} \end{align}$$

where $P_m(x)$ is a polynomial of order $2m$.

The result of this has actually increased the difficulty in evaluating the limit of interest.

So, let's pursue alternative ways forward.

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Since we can represent $e^x$ by its Taylors series, $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$,then clearly for $x>0$ and any integer $|m|$, $e^x\ge \frac{x^{|m|+1}}{(m+1)!}$. Therefore, we see that

$$\begin{align} \left|x^m e^{-1/x}\right|&=\left|\frac{x^m}{e^{1/x}}\right|\\\\ &\le \left|\frac{x^m}{\frac{(1/x)^{|m|+1}}{(|m|+1)!}}\right|\\\\ &=(m+1)!x^{m+|m|+1}\tag2 \end{align}$$

The right-hand side of $(2)$ approaches $0$ as $x\to 0^+$ and we are done!

Answer 2

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.

Answer 3

take $ln$ first, $ln(\frac{e^{-1/x}}{x^m})=-\frac{1+mxln(x)}{x}$, use L'Hospital, we get $mln(x)+m$ goes to $-\infty$, thus apply exp again, we get zero.