Asymptotes of $\arctan (2x)$

Question

My book tells me the horizontal asymptotes of $\arctan2x$ is either at positive or negative $\frac{\pi}{2}$, yet the vertical asymptotes of $\tan2x$ occurs at positive or negative $x=\frac{\pi}{4}$, so obviously the horizontal asymptotes of $\arctan2x$ should be at either positive or negative $y=\frac{\pi}{4}$.

Can someone tell me what I'm doing wrong?

Answer 1

$$\lim\limits_{x\to \infty} \arctan(2x)=\pi /2\\ \lim\limits_{x\to -\infty} \arctan(2x)=-\pi /2$$

Notice that $\frac 1 2 \tan(\arctan(2x))=x$

Answer 2

The functions $x \mapsto \arctan 2x$ and (some restriction of) $x \mapsto \tan 2x$ are not inverses: The inverse of $x \mapsto \arctan 2x$ is $$x \mapsto \tfrac{1}{2} \tan x, \qquad -\tfrac{\pi}{2} < x < \tfrac{\pi}{2} .$$