Question

${lim}_{x\to 0}\frac{\tan x-x}{x^2\tan x}$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{3}$

We know that

$\tan x=x+\frac{x^3}{3}+\frac{2}{15}{x}^5+...$

Now

$\underset{x\to 0}{lim}\frac{\tan x-x}{x^2\tan x}$

$=\underset{x\to 0}{lim}\frac{(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+...)-x}{x^2(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+...)}$

$=\underset{x\to 0}{lim}\frac{\frac{x^3}{3}+\frac{2}{15}{x}^5+...}{x^3+\frac{x^5}{3}+\frac{2x^7}{15}...}$

$=\underset{x\to 0}{lim}\frac{x^3(\frac{1}{3}+\frac{2x^2}{15}+...)}{x^3(1+\frac{x^2}{3}+\frac{2x^4}{15}+...)}$

Cancelling $x^3$ and putting $x=0$, we get-

$=\frac{1}{3}$

Answer-(C)