Question

${lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^3}$ is equal to :-

Answer 1

${lim}_{x\to 0}\frac{2sinx-\sin 2x}{x^3}=\frac{0}{0}$ form

Applying $L-$ Hospital rule

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

Again applying $L-$ Hospital's rule

${lim}_{x\to 0}\frac{-2sinx+4\sin 2x}{6x}=\frac{0}{0}$ form

Applying $L-$ Hospital's rule

${lim}_{x\to 0}\frac{-2cosx+8\cos 2x}{6}=\frac{-2+8}{6}=\frac{6}{6}=1$