Question

Value of $\displaystyle \lim_{x \to 0} \frac{x\,tan \,2x - 2x\, tan \,x}{\left(1-cos \,2x\right)^{2}}$ is

• A. 2
• B. -2
• C. 1/2
• D. -1/2

Answer 1

Answer: (C)

Consider $\displaystyle \lim_{x \to 0} \frac{x\,tan \,2x - 2x\, tan \,x}{\left(1-cos \,2x\right)^{2}}$
$= \displaystyle \lim_{x \to 0} \frac{x\,tan \,2x - 2x\, tan \,x}{4\,sin^{4} \,x}$
$\left(\because cos \,2x = 1 - 2 \,sin^{2}x\right)$
$= \displaystyle \lim_{x \to 0} \frac{x}{4\,sin^{4} \,x} \left[\frac{2\, tan\, x}{1 - tan^{2} \,x} - 2\, tan\, x\right]$
$= \displaystyle \lim_{x \to 0} \frac{2x \,tan\, x}{4\,sin^{4} \,x} \left[\frac{1-1+tan^{2} \,x}{1-tan^{2} \,x}\right]$
$= \displaystyle \lim_{x \to 0} \frac{2x \,tan^{3} \,x}{4\,sin^{4} \,x \left(1-tan^{2}\, x\right)}$
$= \frac{1}{2} \displaystyle \lim_{x \to 0} \frac{x}{sin \,x}. \frac{1}{cos^{3} \,x}. \frac{1}{1-tan^{2}\, x} = \frac{1}{2}.1. \frac{1}{1^{3}}. \frac{1}{1-0} = \frac{1}{2}$