Proving that for all complex $z$, $\lim_{x\to0}\frac{1-\cos^{z}x}{x^2}=\frac{z}{2}.$

Question

What do I need to study beforehand in order to prove it (not necessarily in only one way)? I found this sperimentally, at the moment we're beginning derivatives at school. By induction, I succeeded in proving $$\lim_{x\to0}\frac{1-\cos^{n}x}{x^2}=\frac{n}{2}.\tag{P(n)}$$

Basis: $P(0)$: $\displaystyle\lim_{x\to0}\frac{1-\cos^{0}x}{x^2}=0$ is true.

Inductive step: Assume $P(k)$ holds. This indeed implies $P(k+1)$ holds too: $$\lim_{x\to0}\frac{1-\cos^{k+1}x}{x^2}=\lim_{x\to0}\frac{\left(1-\cos^kx\right)\cos x-\cos x+1}{x^2}=\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lim_{x\to0}\frac{\left(1-\cos^kx\right)\cos x}{x^2}+\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{k}{2}+\frac{1}{2}=\frac{k+1}{2}.\ \ \ \ \ \ \ \ \ \ \ \ \ \blacksquare$$

I want to show it is true for all complex $z$ on my own.

Answer 1

$$\frac{1-\cos^z x}{x^2}=\frac{1-\exp(z\log \cos x)}{x^2}=\frac{1-\exp(z\log \cos x)}{\log\cos x}\cdot\frac{\log\cos x}{x^2}$$ But since $\log\cos x\to 0$ when $x\to 0$, $$\lim_{x\to 0}\frac{1-\cos^z x}{x^2}=-z\cdot\lim_{x\to 0}\frac{\log\cos x}{x^2}\stackrel{H}{=}z\cdot\lim_{x\to 0}\frac{\tan x}{2x}=\frac{z}{2}.$$

Answer 2

What about a little l'Hospital?

$$\lim_{x\to 0}\frac{1-\cos^zx}{x^2}=\frac{z\cos^{z-1}x}2\frac{\sin x}x\xrightarrow[x\to 0]{}\frac z2\cdot1=\frac z2$$