Question

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

• A. $\frac{1}{2}$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}=\underset{x\to 0}{lim}\frac{\frac{\sin x}{\cos x}-\sin x}{x^3}$
$=\underset{x\to 0}{lim}-\frac{\sin x(\cos x-1)}{x^3\cos x}=\underset{x\to 0}{lim}\frac{1}{\cos x}\times \underset{x\to 0}{lim}-\frac{\sin x(\cos x-1)}{x^3}$
$=\underset{x\to 0}{lim}-\frac{\sin x(\cos x-1)}{x^3}$
Applying L Hospital's rule
$=-\underset{x\to 0}{lim}\frac{-\cos x+{\left(\cos x\right)}^2-{\left(\sin x\right)}^2}{3x^2}$
Applying L Hospital's rule
$=-\underset{x\to 0}{lim}\frac{\sin x-4\cos x\sin x}{6x}$
Again applying L Hospital's rule
$=-\frac{1}{6}\underset{x\to 0}{lim}(\cos x-4{\left(\cos x\right)}^2+4{\left(\sin x\right)}^2)$
$=-\frac{1-4+0}{6}=\frac{1}{2}$