Proof $ \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos^{2}x} - 4 = \infty $

Question

I want to prove that $$ \lim_{x \to \frac{\pi}{2}} \left(\frac{1}{\cos^{2}x} - 4\right) = \infty ,$$ is my proof correct?

Proof:

Given $ M \ge 1$, choose $ \delta = \arccos\left(\sqrt{\frac{1}{M}}\right) - \frac{\pi}{2}$.

Suppose $ 0 \lt \left|x - \frac{\pi}{2}\right| \lt \delta $.

Therefore:

$ x - \frac{\pi}{2} \lt \arccos(\sqrt{\frac{1}{M}}) - \frac{\pi}{2} $

$ x \lt \arccos(\sqrt{\frac{1}{M}}) $

$ \cos\left(x\right) \lt \sqrt{\frac{1}{M}}$

$ \cos^{2}\left(x\right) \lt \frac{1}{M}$

$ \frac{1}{\cos^{2}\left(x\right)} \gt M $

$ \frac{1}{\cos^{2}\left(x\right)} - 4 \gt M $

Answer 1

Given $M > 0$ we solve $\vert\frac{1}{cos^2x} - 4\vert \geqslant M$ with $x \in [0, 2\pi]$ $$\frac{1}{cos^2x} \geqslant M + 4 \quad\vee\quad \frac{1}{cos^2x} \leqslant 4 - M$$ WLOG we assume $M > 4$ $$cos^2x \leqslant \frac{1}{M + 4} \quad\wedge\quad x \neq \frac{\pi}{2}$$ $$-\frac{1}{\sqrt[]{M + 4}} \leqslant cosx \leqslant \frac{1}{\sqrt[]{M + 4}}$$ Set $k = \arccos\left(\frac{1}{\sqrt[]{M + 4}}\right)$ and find $k \leqslant x < \frac{\pi}{2} \quad\vee\quad \frac{\pi}{2} < x \leqslant \pi - k$

So $A = [k, \frac{\pi}{2}) \quad\cup\quad (\frac{\pi}{2}, \pi - k]$ with $0 < k < \frac{\pi}{2}$

A is a neighborhood of $\frac{\pi}{2}$ and $0 < \delta \leqslant \frac{\pi}{2} - k$