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Let $f(x)=e^{\frac{1}{x}}$. In which of the following domains is it uniformly continuous?

a) (0,1)

b) (1,$\infty$)

c) (1,2)

My Attempt: For uniform continuity, we have a given $\epsilon$>0 and we need to find an $\delta$>0 such that whenever $|x-y|<\delta$, the following holds : $|f(x)-f(y)|<\epsilon$. In case (b) and (c), clearly the function $e^{\frac{1}{x}}$ is bounded. So we can always find such an $\delta$, where the above result holds true. But in case (a) when $x$ takes values very close to $0$, the function $f(x)$ diverges. So we can say it is not uniformly continuous.

Is my approach and justification correct? Can someone please give proper reasoning and mathematical explanation because I am not completely satisfied with what I have presented above.

Thanks in advance.

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On intervals where your function has a Lipschitz constant (essentially, a bound on the first derivative) it will be uniformly continuous. Your function has a bounded derivative on both $(1,2)$ and $(1,\infty)$, so it is UC there. It is not UC on $(0,1)$ since you have arbitrarily large slopes close to $0$. In other words you can pick two points close to zero in such a way that $$ \frac{f(x_1)-f(x_2)}{x_1-x_2}\quad (*) $$ is as large as you want. Given an $\varepsilon$, no uniform $\delta$ will satisfy the definition of UC, you would have to choose a smaller and smaller $\delta$ the closer you get to zero, because of $(*)$.

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  • $\begingroup$ I have not yet come across a Lipschitz function. Thanks for the help. +1 $\endgroup$ – s1mple Oct 7 '19 at 2:35
  • $\begingroup$ Can you please say something about my approach? the idea is it correct? etc $\endgroup$ – s1mple Oct 7 '19 at 2:36
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    $\begingroup$ About your approach: being bounded is not enough to be able to choose a universal $\delta$. For example, take $f(x)=\sin(1/x)$ on $(0,1)$. This function is bounded but it is not uniformly continuous because it has oscillations with arbitrarily large frequency close to zero. What you need is to have a controlled slope. Your argument for part $a)$ seems correct: if your function has limit $\pm\infty$ at the end-point, it cannot be UC. $\endgroup$ – GReyes Oct 7 '19 at 3:14
  • $\begingroup$ Thank you for the help. +1 $\endgroup$ – s1mple Oct 7 '19 at 3:16

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