Evaluate $\lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right)$
$\begin{aligned} & \lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right) & \left(\infty-0 \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\log \left(1-x\right)}{\tan \left(\dfrac{\pi x}{2}\right)} & \left(\dfrac{\infty}{\infty} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\log \left(1-x\right)\right]}{\dfrac{d}{dx}\left[\tan \left(\dfrac{\pi x}{2}\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{-1}{1-x}}{\dfrac{\pi}{2}\sec^2 \left(\dfrac{\pi x}{2}\right)} & \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\cos^2 \left(\dfrac{\pi x}{2}\right)}{x-1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\cos^2 \left(\dfrac{\pi x}{2}\right)\right]}{\dfrac{d}{dx}\left(x-1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{-2 \cos \left(\dfrac{\pi x}{2}\right) \times \sin \left(\dfrac{\pi x}{2}\right) \times \dfrac{\pi}{2}}{1} & \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; - \sin \left(\dfrac{2 \pi x}{2}\right) & \left[\text{Note: }\sin 2 \theta = 2 \sin \theta \cos \theta\right] \\ & & \\ & = - \lim\limits_{x \to 1^{-}} \; \sin \left(\pi x\right) & \\ & & \\ & = 0 \end{aligned}$