First slide

Introduction to limits

Question

$lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{x \tan 4 x}$ is equal to

Moderate

A

$- 1 / 4$
B

$1 / 2$
C

1
D

2

Solution

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

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