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. 

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*}
