One-sided infinite limit of cot(x) at pi

Rewrite $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$. As $x \to \pi$, $\cos(x) \to \cos(\pi) = -1$, so the numerator approaches $-1$. Meanwhile $\sin(x) \to \sin(\pi) = 0$, so the denominator approaches $0$, making the fraction unbounded in magnitude.

To determine the sign, take $x$ slightly greater than $\pi$ (e.g. $x = \pi + 0.01$), so $\sin(x)$ is negative and close to $0^{-}$. Dividing the negative numerator $\cos(x)\approx -1$ by a small negative denominator gives a large positive quotient, so $\cot(x) \to +\infty$ as $x \to \pi^{+}$.