$\lim_{x\to 1}\ln(1-x)\cot\frac{\pi x}2$ $$\lim_{x\to 1}\ln(1-x)\cot\frac{\pi x}2$$
After applying L'Hospital twice, I get 
$$\lim_{x\to 1}\frac{-2\sin\pi x}\pi = 0$$
Is this correct? And if I do by LHL and RHL method, ln(1-x) would not be defined for RHL since the log of negative is not defined. Also the concern that tan($\pi$/2) approaches +$\infty$ from LHL and -$\infty$ from RHL. 

$$\lim_{x\to1^-}\frac{\log(1-x)\cos\frac{\pi x}2}{\sin\frac{\pi x}2}=\lim_{x\to1^-}\frac1{\sin\frac{\pi x}2}\cdot\lim_{x\to1^-}\frac{\log(1-x)}{\frac{1}{\cos\frac{\pi x}2}}\stackrel{\text{l'H}}=$$
$$=1\cdot\lim_{x\to1^-}\frac2\pi\frac{-\frac1{1-x}}{-\frac{\sin\frac{\pi x}2}{\cos^2\frac{\pi x}2}}=\frac2\pi\lim_{x\to1^-}\frac1{\sin\frac{x\pi}2}\lim_{x\to1^-}\frac{\cos^2\frac{\pi x}2}{1-x}\stackrel{\text{l'H}}=1\cdot\lim_{x\to1^-}\frac{-2\cos\frac{\pi x}2\sin\frac{\pi x}2}{-1}=0$$
So yes: your result is correct but only as one-sided limit, as the function's isn't defined on any right neighborhood of $\;1\;$ (the splitting of limits is justified since each limit by itself exists finitely)
