Question

The value of $\underset{x\to 0}{lim}\frac{6\sin x+3\sin 2x-4\sin 3x}{x^2\sin x}$ is

Answer 1

$\underset{x\to 0}{lim}\frac{6\sin x+3\sin 2x-4\sin 3x}{x^2\sin x}=\underset{x\to 0}{lim}\frac{6\sin x+6\sin x\cos x-4(3\sin x-4{\sin }^{3}x)}{x^2\sin x}=\underset{x\to 0}{lim}\frac{6+6\cos x-12+16{\sin }^{2}x}{x^2}=\underset{x\to 0}{lim}\frac{6\cos x-6+16{\sin }^{2}x}{x^2}=16\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}-6\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=16-3=13$

$\text{Alternate Solution:}$

$L=\underset{x\to 0}{lim}\frac{6\sin x+3\sin 2x-4\sin 3x}{x^2\sin x}=\underset{x\to 0}{lim}\frac{6(x-\frac{x^3}{3!}+...)+3(2x-\frac{(2x{)}^3}{3!}+...)-4(3x-\frac{(3x{)}^3}{3!}+...)}{x^2(x-\frac{x^3}{3!}+...)}=\underset{x\to 0}{lim}\frac{6x-\frac{6x^3}{3!}+6x-\frac{24x^3}{3!}-12x+\frac{108x^3}{3!}+.....}{x^3(1-\frac{x^2}{3!}+...)}=\frac{-6-24+108}{3!}=13$