Answer
\[\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}\]
Work Step by Step
Let \[l=\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}\]
Which is $\frac{0}{0}$ form
Using L'Hopital's rule
\[l=\lim_{x\rightarrow 0}\frac{(\cos mx-\cos nx)'}{(x^2)'}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{-m\sin mx+n\sin nx}{2x}\]
Which is again a $\frac{0}{0}$ form
Using L'Hopital's rule
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{(-m\sin mx+n\sin nx)'}{(2x)'}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{-m^2\cos mx+n^2\cos nx}{2}\]
\[\Rightarrow l=\frac{n^2-m^2}{2}\]
Hence , \[\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}\]