1
$\begingroup$

let $f(x)=\dfrac{e^{\large \frac{-1}{x}}}{x}$ where $x\in (0,1)$ .then on $(0,1)$

1.$f$ is uniformly continuous

2.$f$ is continuous but not uniformly continuous

3.$f$ is bounded

4.$f$ is not continuous

I found it to be continuous but not uniformly as $\lim_{x\rightarrow 0} f(x)$ does not exist.It is bounded below by 0 and above by 1.Am i right? Would be grateful if someone could suggest me required edits.

$\endgroup$
  • 2
    $\begingroup$ $$\lim_{x\to 0^+}\frac{e^{-1/x}}x\stackrel{y=\frac1x}=\lim_{y\to\infty } \frac y{e^{y}}\stackrel {l'H}=\lim_{y\to\infty}\frac1{e^y}=0$$ so the limit exists (finitely) when $\;x\to 0^+\;$ $\endgroup$ – Timbuc Jan 15 '15 at 16:16
  • 1
    $\begingroup$ thanks @Timbuc for your answer $\endgroup$ – Learnmore Jan 16 '15 at 2:28
0
$\begingroup$

$$\lim_{x\to 0}\frac{e^{-\frac{1}{x}}}{x}=0$$ therefore the function is uniformly continuous because you can prolonge $x\mapsto \frac{e^{-\frac{1}{x}}}{x}$ to $0$ and the prolongement is continuous on $[0,1]$. But you know that a continuous function on a compact is uniformly continuous, therefore it's uniformly continuous on $[0,1]$ and thus on $]0,1[$.

We can then conclude that 1. and 3. are true and 2. and 4. wrong.

$\endgroup$
0
$\begingroup$

f is bounded beaouse limit to 0+ exists, and on (0,1] is continious is we define f(0)=0 than f is continious in [0,1]=> f have max on [0,1], f on compact that is contunious is uniform continious.

$\endgroup$
  • $\begingroup$ This answer, in spite of its being written unclearly and having some tiny gaps (which the OP can complete) in its argumentation, is correct. Why the downvote? +1 $\endgroup$ – Timbuc Jan 16 '15 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.