Question

If $\underset{x\to 0}{lim}\frac{5\sin x-7\sin 2x+3\sin 3x}{x^2\sin x}=a$Find $-a$

Answer 1

$\underset{x\to 0}{lim}\frac{5\sin x-7\sin \left(2x\right)+3\sin \left(3x\right)}{x^2\sin x}$ $\left\{\begin{array}{c}\sin \left(2\theta \right)=2\sin \theta \cos \theta \\ \sin \left(3\theta \right)=3\sin \theta -4{\sin }^3\theta \end{array}\right\}$

$=\underset{x\to 0}{lim}\frac{5\sin x-14\sin x\cos x+9\sin x-12{\sin }^{3}x}{x^2\sin x}$

$=\underset{x\to 0}{lim}\frac{5-14\cos x+9-12{\sin }^{2}x}{x^2}$

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

$=\underset{x\to 0}{lim}\frac{28}{4}\frac{{\sin }^2\left(\frac{x}{2}\right)}{(\frac{x}{2}{)}^2}-\underset{x\to 0}{lim}12{\left(\frac{\sin x}{x}\right)}^2$

$=7-12$

$=-5$.