First slide

Introduction to limits

Question

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

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{4}$
C

$\frac{2}{3}$
D

0

Solution

$\begin{array}{l} lim_{x \to 0} \frac{xtan 2 x - 2 xtan x}{(1 - \cos 2 x)^{2}} \\ = lim_{x \to 0} \frac{x \{2 x + \frac{8 x^{3}}{3} + \frac{64 x^{5}}{15} + \dots\} - 2 x \{x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots\}}{4 \sin^{4} x} \\ = lim_{x \to 0} \frac{x^{4} \{(\frac{8}{3} - \frac{2}{3}) + terms containing positive power of x\}}{4 \sin^{4} x} \\ = \frac{1}{4} \cdot 2 \times \frac{1}{lim_{x \to 0} {(\frac{\sin x}{x})}^{4}} = \frac{1}{2} \times \frac{1}{1} = \frac{1}{2} \end{array}$

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