limx→0 (1−cos⁡2x)22xtan⁡x−xtan⁡2x, is - Infinity Learn by Sri Chaitanya

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

Solution:

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

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

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