Answer
$$\lim_{x\to0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\cos mx-\cos nx}{x^2}$$
For $\lim_{x\to0}(\cos mx-\cos nx)=\cos(0m)-\cos(0n)=\cos0-\cos0=0$ and $\lim_{x\to0}(x^2)=0^2=0,$
we have an indeterminate form of $\frac{0}{0}$, and we can apply L'Hospital's Rule here:
$$A=\lim_{x\to0}\frac{(\cos mx-\cos nx)'}{(x^2)'}$$
$$A=\lim_{x\to0}\frac{-m\sin mx-(-n\sin nx)}{2x}$$
$$A=\lim_{x\to0}\frac{-m\sin mx+n\sin nx}{2x}.$$
$\lim_{x\to0}(-m\sin mx+n\sin nx)=-m\sin(0m)+n\sin(0n)=-m\sin0+n\sin0=-0m+0n=0$
and $\lim_{x\to0}(2x)=2\times0=0,$
which is another indeterminate form of $\frac{0}{0}$. Applying L'Hospital's Rule, we have
$$A=\lim_{x\to0}\frac{(-m\sin mx)'+(n\sin nx)'}{(2x)'}$$
$$A=\lim_{x\to0}\frac{-m^2\cos mx+n^2\cos nx}{2}$$
$$A=\frac{-m^2\cos(0m)+n^2\cos(0n)}{2}$$
$$A=\frac{-m^2\cos0+n^2\cos0}{2}$$
$$A=\frac{-m^2\times1+n^2\times1}{2}$$
$$A=\frac{n^2-m^2}{2}$$
