Question

Solve the equation : ${lim}_{x\to 0}\frac{\tan x-\sin x}{x^2}=?$

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

Answer 1

The correct option is A $0$

$\underset{x\to 0}{lt}\frac{tanx-sinx}{x^2}$

$=\frac{{\sec }^{2}x-cosx}{2x}$

$=\frac{2{\sec }^{2}xtanx+sinx}{2}$

$=0$