Question

Evaluate :-
L=${lim}_{x\to 0}\frac{3\sin x-\sin 3x}{x^3}$

Answer 1

${lim}_{x\to 0}\frac{3\sin x-\sin 3x}{x^3}$

it is of the form $\frac{0}{0}$

So, we will apply L Hospital rule until we will get determinate form

Differentiating numerator and denominator we will get,

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

Differentiating numerator and denominator again we will get,

${lim}_{x\to 0}\frac{-3\sin x+9\sin 3x}{6x}$

Differentiating numerator and denominator again we will get,

${lim}_{x\to 0}\frac{-3\cos x+27\cos 3x}{6}=\frac{-3+27}{6}=4$