Is $\liminf_{n \to \infty} \arctan(x_n) = \arctan( \liminf_{n \to \infty} x_n)$?

Question

Let $\{x_n\}_{n=1}^\infty$ be any sequence of real numbers. In a proof I'm reading, it is used that

$$\liminf_{n \to \infty} \arctan(x_n) = \arctan( \liminf_{n \to \infty} x_n)$$

where $\arctan: [-\infty, + \infty] \to [-\pi/2, \pi/2]$ is the inverse of $\tan$ extended with $\arctan(\pm\infty) := \pm \pi/2$;

It seems odd to me that you can interchange the $\arctan$ and $\liminf$. I can find counterexamples for continuous functions that this does not hold in general. Is there any reason why this relation does hold for $\arctan$? Or is my book wrong?

Answer 1

It is true for any increasing continuous function (on the extended real line with finite limits at $\pm \infty)$. $\arctan $ has these properties.