L'Hôpital's ain't helpin' on this one

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to \infty} x^2 \ln (x \cot^{-1}x) & \small \color{#3D99F6} \text{Let }\theta = \cot^{-1} x \implies x = \frac 1{\tan \theta} \\ & = \lim_{\theta \to 0} \frac {\ln \left(\frac \theta{\tan \theta} \right)}{\tan^2 \theta} \\ & = \lim_{\theta \to 0} \ln \left(\frac \theta{\color{#3D99F6} \tan \theta} \right) \times \frac 1{\color{#3D99F6} \tan^2 \theta} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{\theta \to 0} \ln \left(\frac \theta{\color{#3D99F6} \theta + \frac 13 \theta^3 + \frac 2{15}\theta^5 + \cdots} \right) \times \frac 1{\color{#3D99F6}\left(\theta + \frac 13 \theta^3 + \frac 2{15}\theta^5 + \cdots\right)^2} \\ & = \lim_{\theta \to 0} - \frac {\color{#3D99F6} \ln \left(1 + \frac 13 \theta^2 + \frac 2{15}\theta^4 + \cdots \right)}{\left(\theta + \frac 13 \theta^3 + \frac 2{15}\theta^5 + \cdots\right)^2} & \small \color{#3D99F6} \text{By Maclaurin series again} \\ & = \lim_{\theta \to 0} - \frac {\color{#3D99F6} \frac 13 \theta^2 + O(\theta^4)}{\theta^2+O(\theta^4)} & \small \color{#3D99F6} \text{Divide up and down by }\theta^2 \\ & = \lim_{\theta \to 0} - \frac {\frac 13 + O(\theta^2)}{1+O(\theta^2)} \\ & = \boxed{-\dfrac 13} \end{aligned}$