In order to prove that $\displaystyle\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$, with $a \ne 0$, I managed that $a=2$ and evaluated this limit:
$ \begin{align*} \quad \lim_{x \to 0}\frac{1-\cos(2x)}{2x}&= \lim_{x \to 0}\frac{1-(1-2\sin^2(x))}{2x}\\ &= \lim_{x \to 0}\frac{1-1+2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{\sin^2(x)}{x}\\ &= \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \sin(x)\\ &= 1 \cdot 0\\ &=0 \end{align*}$
Can I generalize it?