Answer
$1$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to \pi}(x-\pi) \cot x=\lim\limits_{x \to \pi} \dfrac{(x-\pi)}{\tan (x)}$
But $\lim\limits_{x \to \pi} \dfrac{(x-\pi)}{\tan (x)}=\dfrac{0}{0}$
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.
$\lim\limits_{x \to \pi} \dfrac{(x-\pi)}{\tan (x)}=\lim\limits_{x \to \pi} \dfrac{1}{\sec^2 x}\\= \dfrac{1}{\sec^2 (\pi)}\\=1$
