# find Asymptotes of $f(x) = \arcsin(\frac{2x}{1+x^2})$

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

1. 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.

2. 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:
You can show that the limit of $\arcsin\left(\frac{2x}{1+x^2}\right)=0$ for both $x\to +\infty$ and $x \to -\infty$. Let's first find the limit at $x \to +\infty$: $$\lim_{x\to +\infty} \arcsin\left(\frac{2x}{1+x^2}\right)=\lim_{x\to +\infty} \arcsin\left(\frac{2}{\frac{1}{x}+x}\right)$$ $\frac{1}{x}$ tends to $0$ as $x\to \infty$ and $x$ tends to $\infty$. Therefore: $$\frac{2}{\frac{1}{x}+x}\to 0$$ And $\arcsin(0)=0$. Therefore: $$\lim_{x\to +\infty} \arcsin\left(\frac{2x}{1+x^2}\right)=0$$ You may similarly show that: $$\lim_{x\to -\infty} \arcsin\left(\frac{2x}{1+x^2}\right)=0$$ The horizontal asymptotes are given by $y=b$ where $b$ are the limits where $x \to \pm \infty$. Therefore, since both values of $b=0$, the horizontal asymptote is $y=0$.