Answer
$-\frac{2}{\pi}.$
Work Step by Step
We have
$$
\lim _{x \rightarrow 1}\tan (\frac{\pi x}{2})\ln x=\lim _{x \rightarrow 1}\frac{\sin(\frac{\pi x}{2})\ln x}{\cos\frac{\pi x}{2} }=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 1}\frac{\sin(\frac{\pi x}{2})\ln x}{\cos\frac{\pi x}{2} }= \lim _{x \rightarrow 1}\frac{\frac{1}{x}\sin(\frac{\pi x}{2})+\frac{\pi}{2}\cos(\frac{\pi x}{2}) \ln x}{-\frac{\pi}{2}\sin\frac{\pi x}{2} }\\
=-\frac{2}{\pi}.
$$
