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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Limits

The value of

lim_{x \to 0} (\frac{x}{8 1 - s i n x - 8 1 + s i n x})

is equal to :

(A)4
(B)0
(C)-4✓ Correct
(D)-1

Answer: (C)

Solution

Put

t = sin x

, so

t \to 0

as

x \to 0

and

\frac{x}{t} \to 1

. The limit becomes

t \to 0 lim \frac{t}{( 1 - t ) ^{1/8} - ( 1 + t ) ^{1/8}}

The binomial expansions

(1 - t)^{1/8} = 1 - \frac{t}{8} + O (t^{2})

and

(1 + t)^{1/8} = 1 + \frac{t}{8} + O (t^{2})

make the denominator

- \frac{t}{4} + O (t^{2})

, so the quotient tends to

\frac{t}{- t /4} = - 4

. The limit is

- 4

.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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