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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.

Bernard
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RM777
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  • Sorry, it seems I misunderstood what you meant. I've put it in the previous state. Differentiating a negative number of tiles means calculating the indefinite integral $m$ times, I suppose? – Bernard Feb 11 '19 at 21:17
  • I think it does but I am not sure because we are not there yet in our lecture. But in this case I want to really differentiate because I am applying Hospital. And because I have got $x^{-m}$ where $-m$ is a positive number if $m<0$. I want to take the derivative so often until I got $x^0 \cdot c$ where $c$ is a constant ($(-m)!$) – RM777 Feb 11 '19 at 21:21
  • But the simple substitution $t=\frac1x$ and high-school limits are enough to justify the limit! – Bernard Feb 11 '19 at 21:29

3 Answers3

1

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

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.

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!

Mark Viola
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  • I had an idea yesterday, do you think it is Right? : We have $m<0$ and we have the term $x^me^{-\frac{1}{x}}$. I can rewrite $x^m$ to $\frac{1}{x^{-m}}$. To avoid confusion I define $-m=n$ which is positive. Then I have $x^me^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x^n}=\frac{\sum_{k=0}^{\infty}\frac{(\frac{-1}{x})^k}{k!}}{x^n}=\sum_{k=0}^{\infty}\frac{(\frac{-1}{x^{n+1}})^k}{k!}=e^{-\frac{1}{x^{n+1}}}\rightarrow 0$ as $x\rightarrow 0$ – RM777 Feb 12 '19 at 10:08
  • Not as $x\rightarrow 0$ but $x\downarrow 0$ – RM777 Feb 12 '19 at 10:16
  • There is an error. Note that $x^{-k}/x^n=x^{-(k+n)}$. – Mark Viola Feb 12 '19 at 14:50
  • Sorry for responding so late. I was first confused when you mentioned Taylorseries, we didn't have that yet but we ha ve defined $e^x$ with that powerseries you wrote. Okey I am trying to explain in my own word what you did: $e^{\frac{1}{x}}=\sum_{n=0}^\infty \frac{(\frac{1}{x})^n}{n!}= 1+ \frac{\frac{1}{x}}{1}+\frac{\frac{1}{x^2}}{2!}+\frac{\frac{1}{x^3}}{3!} … + \frac{\frac{1}{x^{|m|+1}}}{(|m|+1)!} + ... \overset{x>0}{\Rightarrow} e^{\frac{1}{x}}>\frac{\frac{1}{x^{|m|+1}}}{(|m|+1)!}.$ We want to prove $\lim_{x\downarrow 0}x^me^{-\frac{1}{x}}=0$ To do so we have to Show : – RM777 Feb 13 '19 at 10:08
  • $\forall_{\epsilon>0}\exists_{\delta>0}\forall_{x>0}|x-0|=|x|=x<\delta \Rightarrow |\frac{x^m}{e^{1/x}}|=\frac{x^m}{e^{1/x}}<\epsilon$. I can ommit the absolute value signs because the values are always psoitive. By the property you proved ago I can say $\frac{x^m}{e^{\frac{1}{x}}}\overset{a>b\Rightarrow \frac{1}{a}<\frac{1}{b}}{<}\frac{x^m}{\frac{\frac{1}{x^{|m|+1}}}{(|m|+1)!}}=\frac{x^m}{\frac{1}{x^{|m|+1}\cdot (|m|+1)!}}$ BUT this must be $\frac{x^m}{1}\cdot \frac{1}{x^{|m|+1}(|m|+1)}\overset{\frac{3^7}{3^2}=3^{7-2}=3^5}{=}\frac{x^{m-(|m|+1)}}{(|m|+1)!} $Isn't that different from your res.? – RM777 Feb 13 '19 at 10:42
  • I figured it out $\frac{x^m}{\frac{\frac{1}{x^{|m|+1}}}{(|m|+1)!}}\overset{\frac{a}{\frac{b}{c}}=\frac{a\cdot c}{b}}{=}\frac{x^m\cdot(|m|+1)!}{\frac{1}{x^{|m|+1}}}\overset{\frac{1}{x^m}=x^{-m}}{=}\frac{x^m\cdot(|m|+1)!}{x^{-(|m|+1)}}^\overset{\frac{x^n}{x^m}=x^{n-m}}{=}x^{m-(-(|m|+1))}(|m|+1)!$ If $m<0$ then $x^{m-(-(|m|+1))}(|m|+1)!=x(|m|+1)!$For$x<(|m|+1)!\epsilon$ we get $x^me^{-\frac{1}{x}}=\frac{x^m}{e^{1/x}}\leq (|m|+1)!x<\epsilon$. For $m\geq 0 \lim_{x\downarrow 0}x^m \geq 1\wedge \lim_{x\downarrow 0} e^{-\frac{1}{x}}=0\Rightarrow \lim_{x\downarrow 0}x^me^{\frac{-1}{x}} =0$. Thank you. – RM777 Feb 13 '19 at 12:15
  • You're welcome. My pleasure. – Mark Viola Feb 13 '19 at 13:36
0

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.

J.G.
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0

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.

Alan Zhang
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