\underset{x \rightarrow 0}{Lim} \textrm{ } \frac{\left(\right. 1 - cos 2 x \left.\right) \textrm{ } \left(\right. 3 + cosx \left.\right)}{x \textrm{ } tan 4 x} is equal to :

Engineering

Mathematics

Trigonometric Limits

Question

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

$\frac{- 1}{4}$

$\frac{1}{2}$

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$\underset{x \to 0}{Lim} \frac{(1 - \cos 2 x)}{{(2 x)}^{2}} {(2 x)}^{2} \cdot (3 + cosx) \cdot \frac{1}{x (\frac{\tan 4 x}{4 x}) \cdot 4 x} = \frac{1}{2} \cdot (3 + 1) \cdot \frac{1}{1} = 2$

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