Evaluating $\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$

The identity given in your question is useful for evaluating the limit: as $x\to 0$ we have $$\frac{\cos(ax) - \cos(bx)}{\sin^2(x)} = -2\frac{\frac{\sin\frac{(ax-bx)}{2}}{x}\cdot \frac{\sin\frac{(ax+bx)}{2}}{x}}{\frac{\sin x}{x}\cdot\frac{\sin x}{x}}\to -2\frac{\frac{a-b}{2}\cdot \frac{a+b}{2}}{1\cdot 1}=\frac{b^2-a^2}{2}$$ where we used the fact that for any real $\alpha$, $$\lim_{x\to 0}\frac{\sin(\alpha x)}{x}=\alpha\lim_{x\to 0}\frac{\sin(\alpha x)}{(\alpha x)}=\alpha\cdot 1=\alpha.$$

\begin{align}\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{\sin^2x}&=\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\frac{b^2-a^2}2\end{align}since$$\cos(ax)-\cos(bx)=-\frac12(a^2-b^2)x^2+o(x^2).$$

$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}=\lim_{x\rightarrow 0}\frac{x^2}{\sin^2(x)}\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{x^2}\\=-\lim_{x\rightarrow 0}\frac{1- \cos(ax)}{x^2}+\lim_{x\rightarrow 0}\frac{1- \cos(bx)}{x^2}=\frac{ b^2-a^2}{2}$$

Given that $$\lim_{x\rightarrow 0}\frac{1- \cos(x)}{x^2}=1/2$$ $$\lim_{x\rightarrow 0}\frac{ \sin(x)}{x}=1. $$