Question

solve
${lim}_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$

Answer 1

$\underset{x\to 0}{lt}\frac{\cos 3x-\cos x}{x^2}$

It is of the form $\frac{1-1}{0}=\frac{0}{0}$

$\therefore$ Applying L'Hospital Rule, we get

$\underset{x\to 0}{lt}\frac{\sin x-3\sin 3x}{2x}$

Again applying L'Hospital Rule,

$\underset{x\to 0}{lt}\frac{\cos x-9\cos 3x}{2}$

$=\frac{\cos 0-9\cos 0}{2}$

$=\frac{1-9}{2}$

$=\frac{-8}{2}$

$=-4$.