$\lim_{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)}$ is equal to:

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Limits

lim_{x \to 0} \frac{x c o t ( 4 x )}{s i n ^{2} x c o t ^{2} ( 2 x )}

is equal to:

Answer: (D)

Write the expression with sines and cosines,

\frac{x c o t 4 x}{s i n ^{2} x c o t ^{2} 2 x} = \frac{x c o s 4 x}{s i n 4 x} \cdot \frac{s i n ^{2} 2 x}{s i n ^{2} x c o s ^{2} 2 x}

. Using

sin 2 x = 2 sin x cos x

\frac{s i n ^{2} 2 x}{s i n ^{2} x} = 4 cos^{2} x

, so it becomes

\frac{x}{s i n 4 x} \cdot \frac{c o s 4 x}{c o s ^{2} 2 x} \cdot 4 cos^{2} x

. As

x \to 0

\frac{x}{s i n 4 x} \to \frac{1}{4}

while the remaining factors tend to

1

1

and

4

, so the limit is

\frac{1}{4} \times 4 = 1

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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