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

JEE Main · 2015 · 04 Apr 2015 · Offline · Limits

lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}

is equal to :

Answer: (B)

Replace each factor by its small-angle size as

x \to 0

1 - cos 2 x = 2 sin^{2} x \approx 2 x^{2}

, the factor

3 + cos x \to 4

, and

tan 4 x \approx 4 x

x tan 4 x \approx 4 x^{2}

. The limit is therefore

\frac{2 x ^{2} \cdot 4}{4 x ^{2}} = 2

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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