Question

Solve it:-
$\underset{x\to 0}{\mathrm{lt}}\frac{5\sin x-7\sin 2x+3\sin 3x}{x^2\sin x}=$

Answer 1

${lim}_{x\to 0}\frac{5\sin x-7\sin 2x+3\sin 3x}{x^2\sin x}$

We get $\frac{0}{0}$ form

Hence by using L'Hospitals rule, we have

$={lim}_{x\to 0}\frac{5\cos x-14\cos 2x+9\cos 3x}{x^2\cos x+2x\sin x}$

We get $\frac{0}{0}$ form

$={lim}_{x\to 0}\frac{-5\sin x+28\sin 2x-27\sin 3x}{-x^2\sin x+2x\cos x+2x\cos x+2\sin x}$

We get $\frac{0}{0}$ form

$={lim}_{x\to 0}\frac{-5\cos x+56\cos 2x-81\cos 3x}{-x^2\cos x-2x\sin x+2\cos x-2x\sin x+2\cos x-2x\sin x+2\cos x}$

$={lim}_{x\to 0}\frac{-5\cos x+56\cos 2x-81\cos 3x}{-x^2\cos x-6x\sin x+6\cos x}$

$=\frac{-5+56-81}{0-0+6}=\frac{-30}{6}=-5$