Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

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

We have,
$\underset{x\to 0}{lim}\frac{tanx-x}{x^{2}tanx}$ $\left[\frac{0}{0}form\right]$
$=\underset{x\to 0}{lim}\frac{se{c}^{2}x-1}{2xtanx+x^{2}se{c}^{2}x}$ [By L' Hospital's Rule]
$\underset{x\to 0}{lim}\frac{2se{c}^{2}xtanx}{2tanx+4xse{c}^{2}x+2x^{2}se{c}^{2}xtanx}$ [By L-Hospita's Rule]
$\underset{x\to 0}{lim}\frac{2se{c}^{4}x+4se{c}^{2}xta{n}^{2}x}{6se{c}^{2}x+12xse{c}^{2}xtanx+2x^{2}se{c}^{4}x+4x^{2}se{c}^{2}xta{n}^{2}x}$
$=\frac{2}{6}=\frac{1}{3}$
Hence, option 'C' is correct.