TITLE: What is $\lim\limits_{z \to \pi/2} \tan^2(z) $ for $z \in \mathbb C$?
QUESTION [6 upvotes]: I am trying to evaluate the following limit ($z \in \mathbb C$):
$$\lim\limits_{z \to \pi/2} \tan^2(z) $$
I get the following solution:
$$\lim\limits_{x \to \pi/2} \tan^2(x) = \lim\limits_{x \to \pi/2} \frac{\sin^2(x)}{\cos^2(x)} = \lim\limits_{x \to \pi/2} \frac{1 - \cos^2(x)}{1-\sin^2(x)} = \infty$$
whereas:
$$\lim\limits_{x \to \pi/2} \tan^2(ix) = \lim\limits_{x \to \pi/2} \frac{\sin^2(ix)}{\cos^2(ix)} = \lim\limits_{x \to \pi/2} \frac{(\frac{1}{2i}(e^{-x} + e^{x}))^2}{(\frac{1}{2}(e^{-x} - e^{x}))^2} = \lim\limits_{x \to \pi/2} \frac{-(e^{-x} + e^{x})^2}{(e^{-x} - e^{x})^2} < \infty$$
Thus, I conclude the limit is undefined. Is that correct?

REPLY [4 votes]: No, it is not correct.
$$\lim_{x\to\pi/2} \tan^2(ix)$$
has no bearing on the actual question. In that limit, you are approaching an argument of $i\pi/2$. You probably meant to consider
$$\lim_{x\to 0}\tan^2\left(\frac{\pi}{2}+ix\right)$$
which is infinite.