First slide

Introduction to limits

Question

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

Moderate

A

2
B

-2
C

$\frac{1}{2}$
D

$\frac{- 1}{2}$

Solution

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

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