find Asymptotes of $f(x) = \arcsin(\frac{2x}{1+x^2})$ I'm trying to find the asymptotes of $f(x) = \arcsin(\frac{2x}{1+x^2})$.


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*I've found that this function has no vertical asymptote, since $f$ is bounded between $[-\pi/2 , \pi/2 ]$, and since $\arcsin x$ is continuous where it is defined - for every $x_0 \in R$, $\lim_{x\to x0^+}|f(x)| = |f(x_0)| \neq \infty $. Hopefully this once is correct, please correct me if it isn't.

*I think I'm wrong in the calculation of the horizontal asymptotes : if $y=ax+b$ is a horizontal asymptote at $\infty$, then $a = \lim_{x\to\infty}\frac{f(x)}{x} = 0$. Now, $b= \lim_{x\to\infty}(f(x)-ax) = \lim_{x\to\infty}f(x) = 0$
So I'm getting that this function has no vertical asymptotes, which I guess is correct, but I also get $y=0$ as a horizontal asymptotes which I'm pretty sure is wrong.. Where is my mistake?

Horizontal asymptotes are of the form $y=\text{constant}$. Use the form $a\,x+b$ for oblique asymptotes.  And there is no mistake. Here is the plot of he function:

