limx→0 xtan⁡2x−2xtan⁡x(1−cos⁡2x)2, is

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

Solution:

$lim_{x \to 0} \frac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^{2}}$

$\begin{array}{l} = lim_{x \to 0} \frac{x (\tan 2 x - 2 \tan x)}{(1 - \cos 2 x)^{2}} \\ = lim_{x \to 0} \frac{2 x \tan^{3} x}{(1 - \tan^{2} x) (1 - \cos 2 x)^{2}} \\ = lim_{x \to 0} \frac{2 {(\frac{\tan x}{x})}^{3}}{(1 - \tan^{2} x) {(\frac{1 - \cos 2 x}{x^{2}})}^{2}} \\ = \frac{2 \times 1^{3}}{(1 - 0) \times 4} = \frac{1}{2} \end{array}$

$lim_{x \to 0} \frac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^{2}}$ , is

Solution:

Related content