Prove that $f : \mathbb R^+ \to\mathbb R$ given by $f(x) = \frac{\ln(1 + x)}x$ is uniformly continuous.

Question

I have been stuck on this for a while. Any help will be appreciated.

Hint: $\lim_{x \to 0} f(x)=1$ and $\lim_{x \to \infty} f(x)=0$. — Kavi Rama Murthy, Nov 08 '19 at 09:44

score 0 · Answer 1 · answered Nov 08 '19 at 09:54

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Use the result that if $f$ is a continuous function on $[0,\infty)$ such that $\operatorname{lim}_{x\to \infty}f(x)$ exists finitely,then $f$ is uniformly continuous.Break the interval $[0,\infty)$ into $[0,a]$ and $[a-\delta_0,\infty)$ suitably and prove that $f$ is uniformly continuous on $[0,\infty)$,then it is uniformly continuous on a subset $(0,\infty)$ also.Or you can use the criterion that if a continuous function has both end point limits finite in an interval of the form $(a,b)$ where $a,b$ are in $[-\infty,+\infty]$,then $f$ is uniformly continuous.

answered Nov 08 '19 at 09:54

Kishalay Sarkar

4,238

@lolol is the proof OK for you? – Kishalay Sarkar Nov 08 '19 at 11:12
I would love it if you could elaborate a bit more – lolol Nov 08 '19 at 13:03
@lolol please see https://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuous I think you will find your doubts cleared here. – Kishalay Sarkar Nov 09 '19 at 04:55

Prove that $f : \mathbb R^+ \to\mathbb R$ given by $f(x) = \frac{\ln(1 + x)}x$ is uniformly continuous.

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