I need to evaluate :
$\lim_{x\to 0} \frac{1-\cos(x)}{x^2}$
In terms of the number:
$\alpha = \lim_{x\to0} \frac{\sin(x)}{x}$
What I am thinking is that $\cos(x) = \frac{\sin(2x)}{2\sin(x)}$
But I don't know if this is the correct beginning
One example (I am doing this just to show that I can't derivate) I already did was:
$\lim_{x\to0} \frac{\sin(ax)}{\sin(bx)} =\lim_{x\to0} \frac{\sin(ax)\cdot bx\cdot a}{ax\cdot \sin(bx)\cdot b} = \alpha \cdot \frac{1}{\alpha}\cdot\frac{a}{b} = \frac{a}{b}$
Hint: $$ 1-\cos x=2\sin^2\frac{x}{2}\Rightarrow \lim_{x\to0} \frac{1-\cos(x)}{x^2}=\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{x^2}=? $$