Evaluate $\underset{x→0}{\lim}\frac{5\sin x-7\sin 2x+3\sin 3x}{x^2\sin x}$.

-5

$\underset{x→0}{\lim}\frac{5\sin x-7\sin 2x+3\sin 3x}{x^2\sin x}=\underset{x→0}{\lim}\frac{5\sin x-7(2\sin x\cos x)+3(3\sin x-4\sin^3x)}{x^2\sin x}$

$=\underset{x→0}{\lim}\frac{14-14\cos x-12\sin^2x}{x^2}=\underset{x→0}{\lim}\frac{14(1-\cos x)-12\sin^2x}{x^2}=\underset{x→0}{\lim}\frac{14(2\sin^2\frac{x}{2})-12\sin^2x}{x^2}$

$=\underset{x→0}{\lim}\frac{28\sin^2x/2}{x^2}\frac{-12\sin^2x}{x^2}=\underset{x→0}{\lim}\frac{7\sin^2x/2}{x^2/4}-12=7-12=-5$