# Proving $f(x)=(\frac {\sin x} {x})^{\frac {1} {x^2} }$ is uniformly continuous on $(0,1]$

Prove that $f(x)=\Large(\frac {\sin x} {x})^{\frac {1} {x^2} }$ is uniformly continuous on $(0,1]$.

Basically what I need to show here is that there is a limit 'from the right' for $x=0$ so the interval can be extended to $[0,1]$ then, any continuous function on a closed interval is uniformly continuous and we're done.

So $\displaystyle\lim_{x\to 0^+}f(x)=...=e^{-\frac 1 6}$ so the extended function would be $g(x)=\begin {cases} f(x) &, x\in (0,1] \\ e^{-\frac 1 6} &,x=0 \end{cases}$ and since $g$ is continuous on a closed interval it is uniformly continuous.

Is it alright ?

At first I tried to show that the derivative of $f$ is bounded but the derivative isn't simple at all and also, even if it's bounded, what about $0$ ? I would still have to extend the function right ?

• Assuming you have no mistakes in the omitted steps of the limit calculation, it is alright, and the best way to prove it that I can see. If you could show the derivative to be bounded, that would be sufficient, it would even prove more than uniform continuity, namely Lipschitz continuity. – Daniel Fischer Jun 21 '14 at 19:11
• @DanielFischer but what about the open part of the interval (for Lipschitz)? – GinKin Jun 21 '14 at 19:11
• Sorry, I don't understand. If $I$ is an interval (open, half-open, closed, finite or infinite, whatever), and $f$ is differentiable on $I$ with bounded derivative, then $f$ is Lipschitz continuous, and a bound for the absolute value of the derivative is a Lipschitz constant. – Daniel Fischer Jun 21 '14 at 19:14
• Oh okay, I wasn't sure if the definition was for any kind of interval. Thanks. – GinKin Jun 21 '14 at 19:15

You are doing it correctly. The limit is not too bad to prove rigorously if you take logs of $f(x)$ and use Taylor series to compute the limit... so you look at $$\lim_{x \rightarrow 0}{ {\ln ({\sin x \over x})} \over x^2}$$ Since ${\sin x \over x} - 1 = - {x^2 \over 6} + O(x^4)$ and $\ln(1 + x) = x +O(x^2)$, the composition $\ln ({\sin x \over x}) = \ln(1 + [{\sin x \over x} - 1])$ equals $-{x^2 \over 6} + O(x^4)$, and the limit is $-{1 \over 6}$. So the right-hand limit at $x = 0$ of $f(x)$ is $e^{-{1 \over 6}}$ as you have there.
• Yeah the limit is fine, I did it with two LHR. The derivative has a lot of room for errors though. How is the limit of $\frac {x^2} 6$ is $\frac 1 6$ as $x\to 0$ ? – GinKin Jun 21 '14 at 19:29
• You are dividing it by $x^2$ there. – Zarrax Jun 21 '14 at 19:30
• I might be confusing things here but isn't that limit arithmetic only allowed as $x\to \infty$ ? – GinKin Jun 21 '14 at 19:31
• This is composing Taylor series and plugging in $x = 0$ basically. Taking $e$ to the limit is just using continuity of $e^x$: $e$ to the limit of $\ln f(x)$ equals the limit of $e^{\ln f(x)}$ – Zarrax Jun 21 '14 at 19:33
• Still why is it allowed to divide by $x^2$ ? – GinKin Jun 21 '14 at 19:50
Maybe the idea is to show that, if $h(x)=f'(x),$ then $h$ may be extended to be continuous on $[0,1]$. This only means determining the limit of $f'(x)$ as $x \to 0,$ which seems numerically to be $0$. [at any $x>0,$ $h$ is continuous by derivative rules.
Anyway, if this is done, then $h$ is continuous on $[0,1]$ and hence bounded there, and then also on $(0,1]$ which is what you need.