Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{\sin 3x-3\sin x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{\sin 3x-3\sin x}$

It becomes 0/0 form.

$=\underset{x\to 0}{lim}\left[\frac{\frac{\sin x}{\cos x}-\sin x}{3\sin x-4{\sin }^{3}x-3\sin x}\right]$

$[\because \sin 3x=3\sin x-4{\sin }^{3}x]$

$=\underset{x\to 0}{lim}\left[\frac{\sin x(1-\cos x)}{\cos x\times (-4{\sin }^{3}x)}\right]$

$=\underset{x\to 0}{lim}\left[\frac{2{\sin }^2\left(\frac{x}{2}\right)}{\cos x\times -4{\sin }^{2}x}\right]$

$[\because 1-\cos x=2{\sin }^2\frac{x}{2}]$

$=\underset{x\to 0}{lim}\left[\frac{2\sin \left(\frac{x}{2}\right)\times \sin \left(\frac{x}{2}\right)}{\cos x\times (-4\sin x)\times \sin x}\right]$

$=\underset{x\to 0}{lim}\left[\frac{2\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\times \frac{x}{2}\times \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\frac{x}{2}}{\cos x\times \frac{-4\sin x}{x}\times x\frac{\sin x}{x}\times x}\right]$

$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=\underset{x\to 0}{lim}\left[\frac{2\times \frac{x}{2}\times \frac{x}{2}}{\cos x\times -4\times x\times x}\right]$

$=\frac{-1}{8\cos 0}$

$=\frac{-1}{8}$