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

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

Solution 1:

$$\lim_{x\to1^-}\frac{\log(1-x)\cos\frac{\pi x}2}{\sin\frac{\pi x}2}=\lim_{x\to1^-}\frac1{\sin\frac{\pi x}2}\cdot\lim_{x\to1^-}\frac{\log(1-x)}{\frac{1}{\cos\frac{\pi x}2}}\stackrel{\text{l'H}}=$$

$$=1\cdot\lim_{x\to1^-}\frac2\pi\frac{-\frac1{1-x}}{-\frac{\sin\frac{\pi x}2}{\cos^2\frac{\pi x}2}}=\frac2\pi\lim_{x\to1^-}\frac1{\sin\frac{x\pi}2}\lim_{x\to1^-}\frac{\cos^2\frac{\pi x}2}{1-x}\stackrel{\text{l'H}}=1\cdot\lim_{x\to1^-}\frac{-2\cos\frac{\pi x}2\sin\frac{\pi x}2}{-1}=0$$

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)

Solution 2:

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 $$ \lim_{t\to0^+}(t\ln t)\frac{\tan t}{t}=0\cdot 1=0 $$