Know your Limits-1

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 1} \frac {\ln(x-1)-x}{\tan \left(\frac \pi {2x} \right)} \quad \quad \small \color{#3D99F6}{\text{This is a }\infty/\infty \text{ case, hence L'Hôpital's rule applies.}} \\ & = \lim_{x \to 1} \frac {\frac 1{x-1}-1}{- \frac \pi{2x^2} \sec^2 \left(\frac \pi {2x} \right)} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down w.r.t }x.} \\ & = \lim_{x \to 1} \frac {2x^2 (x-2) \cos^2 \left(\frac \pi {2x} \right)}{\pi (x-1)} \quad \quad \small \color{#3D99F6}{\text{This is a }0/0 \text{ case, hence L'Hôpital's rule applies.}} \\ & = \lim_{x \to 1} \frac {(6x^2-8x) \cos^2 \left(\frac \pi {2x} \right)+(2x^3-4x^2) \cdot \frac \pi{2x^2} \cos \left(\frac \pi {2x} \right)\sin \left(\frac \pi {2x} \right)}{\pi} \\ & = \boxed{0} \end{aligned}$