Question

$\underset{x\to 0}{lim}\left[\frac{8\sin x+x\sin x}{3\tan x+x^2}\right]$ is equal to

Answer 1

$\underset{x\to 0}{lim}\left[\frac{8\sin x+x\cos x}{3\tan x+x^2}\right]$

$=\underset{x\to 0}{lim}\left[\frac{\frac{8\sin x}{x}+\cos x}{\frac{3\tan x}{x}+x}\right]$

$=\frac{\underset{x\to 0}{lim}[\frac{8\sin x}{x}+\cos x]}{\underset{x\to 0}{lim}[\frac{3\tan x}{x}+x]}$

$=\frac{8+1}{3+0}=\frac{9}{3}=3$