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$$f(x)= {{e^{-{1}\over{x}}}\over {x}}$$ for $x\in (0,1)$ .

Is this function

$a$) uniformly continuous

$b$) bounded but not continuous

$c$) unbounded

This would be uniformly continuous if $f$ could be defined continuously on $0$ and $1$. But that does not look possible.

Help.

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  • $\begingroup$ "This would be uniformly continuous if f could be defined continuously on 0 and 1. But that does not look possible." Why do you say that? $\endgroup$ – zhw. Sep 2 '15 at 15:35
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    $\begingroup$ Substitute $u=1/x$. $\endgroup$ – Urgje Sep 2 '15 at 15:37
  • $\begingroup$ @zhw. Why do I say what ? Why it looks not possible or why it would be uniformly continuous if extended continuously at $0$ and $1$ $?$ $\endgroup$ – user118494 Sep 2 '15 at 15:39
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    $\begingroup$ The exponential tales care of this: $ue^{-u}=u/{e^u}$ $\endgroup$ – Urgje Sep 2 '15 at 16:36
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    $\begingroup$ This is indeed the case. $\endgroup$ – Urgje Sep 3 '15 at 8:07
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Taking clue from urgie's comments :

${e^{-1\over x}}\over x$ is continuous on $(0,1)$ . This will be uniformly continuous if it can be extended continually to $[0,1]$ . Now extending it to $(0,1]$ is no problem . Problem is at the point $0$. Put $u={{1\over x}}$ in the function . Then we have , $$\lim_{x\rightarrow 0}{{e^{-1\over x}}\over x }\\=\lim_{u\rightarrow \infty} u\cdot e^{-u}\\=0$$ as $\lim_{u\rightarrow 0} e^{-u}=0$ and $e^u\gt\gt u$ implies the product with $u$ will certainly go to $0.$ So defining $f(0)=0$ will serve the purpose. Thus $f$ is uniformly continuous.

And of course bounded.

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