Question

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

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

Answer 1

Answer: (D)

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