Is $\frac{1}{\sin x}-\frac{1}{x}$ uniformly continuous on $(0,1)$?

Question

So I am tasked with finding whether $\frac{1}{\sin(x)}-\frac{1}{x}$ is uniformly continuous on the open interval $I=(0,1)$. To look at the "simple" ways to prove it first:

I obviously can't extend the function to $[0,1]$ since the limit at $x=0$ is not defined.

I tried to check if its derivative, $\frac{1}{x^2}-\frac{\cos(x)}{\sin^2(x)}$, is bounded on $I$, in which case the function is uniformly continuous by definition. By using $sin(x)\approx x$ when $x\rightarrow0$ I see that

$\frac{1}{x^2}-\frac{\cos(x)}{\sin^2(x)}\approx \frac{1-\cos(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1-\cos^2(x)}=\frac{1-\cos(x)}{(1-\cos(x))(1+\cos(x))}=\frac{1}{1+\cos(x)}$ when $x\rightarrow0$

So intuitively it looks like it is indeed bounded, but for what $M$ is $|f'(x)|<M$? I haven't shown that it is strictly increasing or decreasing on $I$, so I cant assume $M=max[f'(0),f'(1)]$. Also this method simply feels icky here.

I assume I might have to use epsilon delta to prove/disprove uniform continuity on the interval, but I have simply no idea what values to insert while proceeding with that (the inverse sine bit has me absolutely stumped, I can't seem to relate it to previous assignments using other functions with $sin(x)$ to prove/disprove uniform continuity).

As a side note, I haven't worked with trigonometry in some time so I have a hard time identifying various trigonometric identities off the bat.

Answer 1

As $x>\sin x$ in $(0,1)$ $$f'(x)<{1\over1+\cos x}<1\;\forall\; x\in(0,1)$$ Just replace $\approx$ with $<$. You don't need that $\lim_{x\to0}{\sin x\over x}=1$.

Answer 2

Let's compute $$ \lim_{x\to0}\left(\frac{1}{\sin x}-\frac{1}{x}\right) = \lim_{x\to0}\frac{x-\sin x}{x\sin x} = \lim_{x\to0}\frac{x-x+x^3/6+o(x^3)}{x^2+o(x^2)}=0 $$ Thus the limit does exist.

So your function is the restriction to $(0,1)$ of the continuous function $$ f(x)=\begin{cases} 0 & \text{if $x=0$}\\[6px] \dfrac{1}{\sin x}-\dfrac{1}{x} & \text{if $0<x\le1$} \end{cases} $$ which is uniformly continuous because it is continuous over a closed and bounded interval.

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Note that the limit at $0$ must exist, because a uniformly continuous function on $(0,1)$ extends to a continuous function on $[0,1]$, which is the completion.

The limit can be also computed with l'Hôpital: $$ \lim_{x\to0}\frac{x-\sin x}{x\sin x}= \lim_{x\to0}\frac{x-\sin x}{x^2}= \lim_{x\to0}\frac{1-\cos x}{2x}= \lim_{x\to0}\frac{\sin x}{2}=0 $$ (the first equality exploits the fact that $\lim_{x\to0}\frac{x}{\sin x}=1$).