Question

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

• A. 4
• B. 3
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

We have $ lim_{ x \to 0} \frac{(1 - cos \, 2x) ( 3 + cos \, x)}{x \, tan \, 4x} = lim_{ x \to 0} \frac{ 2 \, sin^2 \, x ( 3 + cos \, x)}{ x \times \frac{tan \, 4x}{ 4x} \times 4 x} $
= $ lim_{ x \to 0} \frac{ 2 \, sin^2 \, x}{ x^2} \times lim_{ x \to 0} \frac{(3 + cos \, x)}{4} \times \frac{1}{ lim_{ x \to 0} \frac{tan \, 4x}{ 4x}} $
= $ 2 \times \frac{4}{4} \times 1 $ $ \Bigg [ \because lim_{ heta \to 0} \frac{ sin \, heta}{ heta} = 1 \, and \, lim_{ heta \to 0 } \frac{tan }{ heta} = 1 \Bigg ]$
= 2.