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

JEE Main · 2013 · 07 Apr 2013 · 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: (C)

Use

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

to recast the top, then lean on the standard small-angle limits. The expression becomes

\frac{2 sin ^{2} x ( 3 + cos x )}{x tan 4 x}

, which groups as

2 (\frac{sin x}{x})^{2} \cdot \frac{4 x}{tan 4 x} \cdot \frac{1}{4} \cdot (3 + cos x)

. As

x \to 0

each ratio tends to

1

and

3 + cos x \to 4

, leaving

2 \cdot 1 \cdot 1 \cdot \frac{1}{4} \cdot 4 = 2

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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