Limit involving inverse functions When I am given the limit
$$\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}$$
would it be possible to evaluate it giving some substitution?
L'Hospital's rule seemed an option but I ended up going in circles. 
 A: When $x>0$, $|x|=x$ and obviously if $x\rightarrow\infty$, then $\sqrt{x^2+1}\rightarrow\infty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $\pi/2$ as its argument goes to infinity:
$$
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2+1}}{\sqrt{x^2+1}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2}\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{|x|\sqrt{1+\frac{1}{x^2}}}=\\
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{x\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{\arctan\sqrt{x^2 +1}}{\sqrt{1+\frac{1}{x^2}}}=
\frac{\pi/2}{\sqrt{1+0}}=\frac{\pi}{2}.
$$
A: You may proceed as follows:


*

*Set $\tan y = \sqrt{1+x^2}$ and consider $y \to \frac{\pi}{2}^-$
\begin{eqnarray*}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}
& = & \sqrt{\tan^2y -1}\frac{y}{\tan y} \\
& = & \frac{\sqrt{\sin^2 y - \cos^2 y}}{\sin y}\cdot y\\
&\stackrel{y \to \frac{\pi}{2}^-}{\longrightarrow} & \frac{\sqrt{1 - 0}}{1}\cdot \frac{\pi}{2} = \frac{\pi}{2}
\end{eqnarray*}
A: Hint: It is true in general that if $\lim f$ and $\lim g$ both exist and are finite and nonzero, then $\lim (fg)$ exists and equals $(\lim f)(\lim g)$.
Take $f(x)=x/\sqrt{x^2+1}$, $g(x)=\tan^{-1}(x^2+1)$ and note that $x\to\infty$ implies $\sqrt{x^2+1}\to\infty$.
