Answer
The solution is
$$\lim_{x\to\pi/2}(\pi-2x)\tan x=2.$$
Work Step by Step
To solve this limit follow the steps below. LR will stand for Apply L'Hopital's rule
$$\lim_{x\to\pi/2}(\pi-2x)\tan x= \lim_{x\to\pi/2}\frac{\pi - 2x}{\cot x}= \left[\frac{\pi-2\frac{\pi}{2}}{\cot\frac{\pi}{2}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\pi/2}\frac{(\pi-2x)'}{(\cot x)'}=\lim_{x\to\pi/2}\frac{-2}{-\frac{1}{\sin^2 x}}=\lim_{x\to\pi/2}2\sin^2 x=2\sin^2\pi/2 = 2\cdot1^2=2.$$
