Calculate $\lim_{x\to \infty}(x + \ln(\frac{\pi}2 - \arctan(x))$ using L'hopital's rule

Question

I'm new to L'hopital's rule. I know i need to convert it to $\frac{\infty}\infty$ or $\frac{0}0$. But I have no idea how to convert the following equation. Thanks in advance for your help!

$$\lim_{x\to \infty}(x + \ln(\frac{\pi}2 - \arctan(x))$$

Answer 1

Hint: Consider $x+\ln\left(\frac{\pi}{2}-\arctan(x)\right)=\\ -\ln(e^{-x})+\ln\left(\frac{\pi}{2}-\arctan(x)\right)=\\ =\ln\left(\frac{\frac{\pi}{2}-\arctan(x)}{e^{-x}}\right)$

Answer 2

You may also proceed as follows using

• $\operatorname{arccot}x = \frac{\pi}{2} - \arctan x$
• $x = \cot u$ while considering the limit for $u \to 0^+$
• $\sin u \ln u = \frac{\sin u}{u}\cdot \underbrace{u \ln u}_{=\frac{\ln u}{\frac{1}{u}}\stackrel{L'Hosp.}{\sim}-u} \stackrel{u \to 0^+}{\longrightarrow} 0$

\begin{eqnarray*} \lim_{x\to \infty}(x + \ln(\frac{\pi}2 - \arctan(x)) & \stackrel{x=\cot u}{=} & \cot u + \ln u \\ & = & \frac{\color{blue}{\overbrace{\cos u + \sin u \ln u}^{\stackrel{u \to 0^+}{\longrightarrow}1}}}{\sin u}\\ & \stackrel{u \to 0^+}{\longrightarrow} & +\infty \end{eqnarray*}