Let $f(x)=\arctan\left({\tan(x/2)\over3}\right)$. Prove that $\lim\limits_{x\to\pi-}f(x)=\pi/4$ and $\lim\limits_{x\to\pi+}f(x)=-\pi/4$

Question

Actually this function is not continuous at $\pi$. Graphically, it can be seen(see graph below) that left hand limit of $f$ at $\pi$ is $\pi/4$ and right hand limit of $f$ at $\pi$ is $-\pi/4$. But how to prove it by hand? It's not obvious to me. Can anybody prove it? Thanks for assistance in advance.

Answer 1

Since $$\lim_{x\to\pi-}{\tan(x/2)\over3}={1\over3}\lim_{x\to\pi/2-}\tan(x)=+\infty$$

so $$\lim_{x\to\pi-}f(x)=\lim_{x\to\pi-}\arctan({\tan(x/2)\over3})=\pi/2$$