I need to check whether the function $f : (0,1) \to \Bbb{R}$ given by $f(x)= \frac{e^{-1/x}}{x}$ is uniformly continuous on its domain.

I know it is continuous since it is a composition of continuous functions. But I have no idea whether it is uniformly continuously or not. I am confused as to how one speculates about whether it will be or not by just looking at it.

I assumed that it is and I tried to estimate $|f(x)-f(y)|$ where $x$ and $y$ are in the domain but I can't proceed. Any help is appreciated. Thanks. :)

  • $\begingroup$ Thanks for the edit. I'm slowly learning MathJax. :) $\endgroup$ – user264750 Nov 25 '15 at 13:42

If you can extend $f$ to $[0,1]$ in such a way that the extension is continuous you are done, since continuos functions on compact sets are uniformly continuous. The extension to $x=1$ is obvious. Can you see how to extend it t0 $x=0$?

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  • $\begingroup$ Oh yes! I know this theorem. I need to find limit of the function as x tends to 0, but I cant find a way to evaluate it. Any hint please? :) $\endgroup$ – user264750 Nov 25 '15 at 13:58
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    $\begingroup$ $$e^{1/x}=1+\frac{1}{x}+\frac{1}{2x^2}+\dots>\frac{1}{2x^2}.$$ Then $$0<\frac{e^{-1/x}}{x}=\frac{1}{x\,e^{1/x}}<2\,x.$$ $\endgroup$ – Julián Aguirre Nov 25 '15 at 14:08
  • $\begingroup$ With due respect, Sir, I don't get how this aids to finding extension at 0. Could you please explain? :) $\endgroup$ – user264750 Nov 25 '15 at 14:19
  • $\begingroup$ What is $\lim_{x\to0}\dfrac{e^{-1/x}}{x}$? $\endgroup$ – Julián Aguirre Nov 25 '15 at 14:22
  • $\begingroup$ 0. I got it. You squeezed the function. Thanks a lot. :) $\endgroup$ – user264750 Nov 25 '15 at 14:30

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