Show derivative function is bounded to prove uniform continuity

Question

I've been trying to solve this question:

Let $f(x) = \frac {x}{x+sin x}$, prove that $f$ is uniformly continuous in $(0,\infty)$

I've been trying to prove that $f'$ is bounded in $(0,\infty)$, but reached:

$f'(x)= \frac {sin x - xcosx}{(x+sin x)^2}$

I'm not sure how to go on from here to show that $f'$ is truly bounded. Would appreciate any help!

Thanks!

Answer 1

For big enough $x$, say $x > A$, we have that $|f'(x)|$ is bounded. Since $f$ has a limit at $0$, we can say that it is a restriction of a continuous function defined on $[0, \infty)$. In particular, $f$ is the restriction of a continous function defined in the compact set $[0, A]$, therefore it's uniformly continuous there and has a bounded derivative at $(A, \infty)$, so it is also uniformly continuous there. Since it is also continuous at $A$ you can argument a bit to show that the original $f$ is uniformly continuous.