Question

Value of $\underset{x\to 0}{\lim }\frac{xtan2x-2xtanx}{{\left(1-cos2x\right)}^2}$ is

• A. 2
• B. -2
• C. 1/2
• D. -1/2

Answer 1

Answer: (C)

Consider $\underset{x\to 0}{\lim }\frac{xtan2x-2xtanx}{{\left(1-cos2x\right)}^2}$
$=\underset{x\to 0}{\lim }\frac{xtan2x-2xtanx}{4si{n}^{4}x}$
$\left(\because cos2x=1-2si{n}^{2}x\right)$
$=\underset{x\to 0}{\lim }\frac{x}{4si{n}^{4}x}\left[\frac{2tanx}{1-ta{n}^{2}x}-2tanx\right]$
$=\underset{x\to 0}{\lim }\frac{2xtanx}{4si{n}^{4}x}\left[\frac{1-1+ta{n}^{2}x}{1-ta{n}^{2}x}\right]$
$=\underset{x\to 0}{\lim }\frac{2xta{n}^{3}x}{4si{n}^{4}x\left(1-ta{n}^{2}x\right)}$
$=\frac{1}{2}\underset{x\to 0}{\lim }\frac{x}{sinx}.\frac{1}{co{s}^{3}x}.\frac{1}{1-ta{n}^{2}x}=\frac{1}{2}.1.\frac{1}{1^3}.\frac{1}{1-0}=\frac{1}{2}$