# 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:

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

You may proceed as follows:

• Set $$\tan y = \sqrt{1+x^2}$$ and consider $$y \to \frac{\pi}{2}^-$$

$$\begin{eqnarray*}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}} & = & \sqrt{\tan^2y -1}\frac{y}{\tan y} \\ & = & \frac{\sqrt{\sin^2 y - \cos^2 y}}{\sin y}\cdot y\\ &\stackrel{y \to \frac{\pi}{2}^-}{\longrightarrow} & \frac{\sqrt{1 - 0}}{1}\cdot \frac{\pi}{2} = \frac{\pi}{2} \end{eqnarray*}$$

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 $$\sqrt{x^2+1}\to\infty$$.