Answer
$-2$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to \pi^{-}} (x-\pi)\tan (\dfrac{x}{2})=\lim\limits_{x \to \pi^{-}} \dfrac{(x-\pi)}{\cot (x/2)}$.
But $\lim\limits_{x \to \pi^{-}} \dfrac{(x-\pi)}{\cot (x/2)}=\dfrac{0}{0}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to \pi^{-}} \dfrac{ \dfrac{d}{dx} (x-\pi)}{ \dfrac{d}{dx}
\cot (x/2)}]=\lim\limits_{x \to \pi^{-}} \dfrac{1}{-\csc^2 (x/2)\times (1/2)} \\ =\dfrac{-2}{\csc^2 (\pi/2)}\\=-2$
