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. 

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:
$$
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2+1}}{\sqrt{x^2+1}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2}\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{|x|\sqrt{1+\frac{1}{x^2}}}=\\
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{x\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{\arctan\sqrt{x^2 +1}}{\sqrt{1+\frac{1}{x^2}}}=
\frac{\pi/2}{\sqrt{1+0}}=\frac{\pi}{2}.
$$
