Answer
$$
\lim _{x \rightarrow 1^{+}} \ln x \tan (\pi x / 2) =-\frac{2}{\pi}
$$
Work Step by Step
$$
\lim _{x \rightarrow 1^{+}} \ln x \tan (\pi x / 2)
$$
We can deal with it by writing the product $ \ln x \tan (\pi x / 2) $ as a quotient $\frac{\ln x}{\cot (\pi x / 2)} $.
Since
$$
\lim _{x \rightarrow 1^{+}} ( \ln x)=0
$$
and
$$
\lim _{x \rightarrow 1^{+} } ( \cot (\pi x / 2))=0
$$
So, we find that, this convert the given limit into an indeterminate form of type $\frac{0}{0}$ and we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 1^{+}} \ln x \tan (\pi x / 2) & =\lim _{x \rightarrow 1^{+}} \frac{\ln x}{\cot (\pi x / 2)} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 1^{+}} \frac{1 / x}{(-\pi / 2) \csc ^{2}(\pi x / 2)} =\frac{1}{(-\pi / 2)(1)^{2}} \\
& =-\frac{2}{\pi}
\end{aligned}
$$
