Question

If the function $f$ is continuous at $x=0$, find $f\left(0\right)$
where $f\left(x\right)=\frac{\cos 3x-\cos x}{x^2},x\ne 0$

Answer 1

Consider the given function.
$f\left(x\right)=\frac{\cos 3x-\cos x}{x^2}$
Since, the function $f\left(x\right)$ is continuous at $x=0$.
Therefore,
${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$
This is the $\frac{0}{0}$ form.
So, apply L-hospital rule,
${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{-3\sin 3x+\sin x}{2x}$
${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\sin x-3\sin 3x}{2x}$
Again apply L-hospital rule, we get
${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\cos x-9\cos 3x}{2}$
On putting the limits, we get
$f\left(0\right)=\frac{1-9}{2}$
$f\left(0\right)=\frac{-8}{2}$
$f\left(0\right)=-4$
Hence, this is the answer.