Question

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

Answer 1

$\underset{x\to 0}{lim}$ $\frac{tanx-sinx}{x^3}$

$=\underset{x\to 0}{lim}$ $\frac{{\sec }^{2}x-cosx}{{3x}^2}$

$=\underset{x\to 0}{lim}$ $\frac{2{\sec }^{2}xtanx+sinx}{6x}$ $=\underset{x\to 0}{lim}$ $\frac{2({\sec }^{2}x+2{\sec }^{2}xtanx)+cosx}{6}=\frac{3}{6}=\frac{1}{2}$