Question

$\lim_{x\to0} \frac{\tan x -\sin x}{x^{3}} = $

• A. $0$
• B. $1$
• C. $ - \frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is D

$\displaystyle\lim_{x\to0} \frac{\tan x -\sin x}{x^{3}} \left(\frac{0}{0} from\right)$
Applying L-Hospital's rule
$ \displaystyle\lim _{x\to 0} \frac{\sec^{2} x -\cos x}{3x^{2}} \left(\frac{0}{0} from\right)$
Again applying L-Hospital's rule,
$ = \displaystyle\lim _{x\to 0} \frac{2 \sec x \sec x \tan x + \sin x}{6x} $
$=\frac{ \displaystyle\lim _{x\to 0} 2 \sec ^{2} x \displaystyle\lim _{x\to 0} \frac{\tan x}{x} + \displaystyle\lim _{x\to 0} \frac{\sin x}{x}}{6}$
$ = \frac{2.1.1+1}{6} = \frac{3}{6} =\frac{1}{2}$