$\lim_{x\to 1}\ln(1-x)\cot\frac{\pi x}2$ $$\lim_{x\to 1}\ln(1-x)\cot\frac{\pi x}2$$
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. 

The function is not defined for $x>1$; in this case writing $\lim_{x\to1}$ or $\lim_{x\to1^-}$ is just a question of conventions.
A possible simplification is to set $1-x=\frac{2t}{\pi}$, so we have
$$
\cot\frac{\pi x}{2}=
\cot\left(\frac{\pi}{2}-t\right)=
\tan t
$$
and the limit becomes
$$
\lim_{t\to0^+}\ln\frac{2t}{\pi}\tan t=
\lim_{t\to0^+}\left(\ln\frac{2}{\pi}\tan t+\ln t\tan t\right)=0
$$
because the first summand has limit $0$ and the second can be written
$$
\lim_{t\to0^+}(t\ln t)\frac{\tan t}{t}=0\cdot 1=0
$$
