Given the function $g(x)=\frac{d}{dx}\left(\frac{x/2}{\sin(x/2)}\right)$, I want to prove it is bounded in $[0,\pi]$.
For me, it is immediate that it is bounded in $(0,\pi]$, as $\frac{x/2}{\sin(x/2)}$ is continuous in $(0,\pi]$. Please, correct me if this needs to be proved with more detail.
Then, how I prove it is bounded in 0? Is it useful to compute the limit of $\frac{x/2}{\sin(x/2)}$ in $0$?. Many thanks in advance!
$$g(x)=f'(x)=\frac{\frac{1}{2}\sin\left(\frac{x}{2}\right)-\frac{x}{4}\cos\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}$$ Simplify $$f'(x)=\frac{\frac{1}{4}\left(2\sin\left(\frac{x}{2}\right)-x\cos\left(\frac{x}{2}\right)\right)}{\sin^2\left(\frac{x}{2}\right)}= \frac{2\sin\left(\frac{x}{2}\right)-x\cos\left(\frac{x}{2}\right)}{4\sin^2\left(\frac{x}{2}\right)}\le\frac{1}{2}$$ From the graph below