Question

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

• A. 1/2
• B. 1
• C. 2
• D. -1/4

Answer 1

Answer: (C)

$\lim_{x\to0}\frac{\left(1 - \cos 2x\right)\left(3 +\cos x\right)}{x\left( \tan 4x\right)}$
$ = \lim _{x\to 0}\frac{1 - \cos 2x}{x^{2}\left(\frac{ \tan 4x}{x}\right)} \left( 3 + \cos x\right) $
$= \lim _{x\to 0}\left(\frac{2\sin^{2}x}{x^{2}}\right). \left(\frac{x}{\tan 4x}\right) \left(3 +\cos x \right)$
$ = 2 \times\frac{1}{4} \times4 = 2$