Question

$\underset{x\to 0}{lim}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)}$ is equal to

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

Answer 1

Answer: (C)

$0$

By L Hospital's Rule,

$=\underset{x\to 0}{lim}\frac{2x({\sec }^{2}2x-{\sec }^{2}x)+\tan 2x-2\tan x}{2sin2x}$

$=\underset{x\to 0}{lim}\frac{2x(4\sec 2x\sec 2x\tan 2x-4\sec x\sec x\tan x)+2({\sec }^{2}2x-{\sec }^{2}x)+2{\sec }^{2}2x-2{\sec }^{2}x}{4\cos 2x}$

$=\frac{0}{1}$

$=0$