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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.

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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:

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

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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*}

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

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