Question

Find the following limit:
$\underset{x\to 0}{lim}\frac{\sin x-\tan x}{{\sin }^{3}x}.$

Answer 1

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

$=\underset{x\to 0}{lim}\frac{\sin x-\frac{\sin x}{\cos x}}{{\sin }^{3}x}=\underset{x\to 0}{lim}\frac{\sin x(1-\frac{1}{\cos x})}{{\sin }^{2}x.\sin x}$

$[\because 1-{\cos }^2\theta =2{\sin }^2\theta \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1]=\underset{x\to 0}{lim}\frac{\cos x-1}{\cos x}\times \frac{1}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{-2{\sin }^2\frac{x}{2}}{\cos x}\times \frac{1}{{\sin }^{2}x}$

$=-2\underset{x\to 0}{lim}\left(\frac{{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}\right)\times \frac{x^2}{4}\times \left(\frac{x^2}{{\sin }^{2}x}\right)\times \frac{1}{x^2}$

$=-2\times \frac{x^2}{4}\times \frac{1}{x^2}$

$=\frac{-2}{4}=\frac{-1}{2}$.