Question

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

• A. 1
• B. \N
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

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

$\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{{\sin }^{3}x}=\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^3}\times \frac{x^3}{{\sin }^{3}x}=\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^3}\times 1=\underset{x\to 0}{\lim }\frac{\left(x+\frac{x^3}{3}+\frac{2x^5}{15}....\right)-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}....\right)}{x^3}=\underset{x\to 0}{\lim }\frac{\left(\frac{x^3}{3}+\frac{2x^5}{15}....\right)-\left(-\frac{x^3}{3!}+\frac{x^5}{5!}....\right)}{x^3}=\underset{x\to 0}{\lim }\frac{x^3\left(\frac{1}{3}+\frac{2x^2}{15}....+\frac{1}{3!}-\frac{x^2}{5!}....\right)}{x^3}=\underset{x\to 0}{\lim }\left(\frac{1}{2}+\frac{x^2}{8}....\right)=\frac{1}{2}$