How to find a one-side limit algebraically of trigonometric function

Question

I was asked to solve one-sided limit of $\arctan(\tan x)$ as $x$ appraoches $\left(\frac{\pi}{2}\right)^+$ and $\left(\frac{\pi}{2}\right)^-$ I tried to sketch a graph but if I'm given for example $\frac{\sin x+\arctan(\tan x)}{x}$ or something like that. How to do it algebraically. I also tried to put a number just a little bit more or less than $\frac{\pi}{2}$ but what should be going then?

Answer 1

Hint: $$ \arctan(\tan x)=x-\pi\left\lfloor\frac{x+\frac{\pi}{2}}{\pi}\right\rfloor $$

Answer 2

In Your second example, the problem comes from $\arctan(\tan(x))$.

We have $$\forall x\in[0,\frac{\pi}{2}) \;\;f(x)=\arctan(\tan(x))=x$$

$$\implies \lim_{x\to \frac{\pi}{2}^-}=\frac{\pi}{2}$$

and

$$\forall x\in (\frac{\pi}{2},\pi)\;\;f(x)=x-\pi$$

$$\implies \lim_{x\to\frac{\pi}{2}^+}=\frac{\pi}{2}-\pi=-\frac{\pi}{2}$$

other approach.

$$\lim_{x\to \frac{ \pi }{2 }^-} \tan(x)=+\infty$$

and

$$\lim_{x\to +\infty}\arctan(x)=+\frac{\pi}{2}$$

$\implies$

$$\lim_{x\to\frac{\pi}{2}^- }\arctan(\tan(x))=+\frac{\pi}{2}$$

the same for $\frac{\pi}{2}^+$.