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JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Limits

If

a = lim_{x \to 0} \frac{1 + 1 + x ^{4} - 2}{x ^{4}}

and

b = lim_{x \to 0} \frac{s i n ^{2} x}{2 - 1 + c o s x}

, then the value of

a b^{3}

is :

(A)36
(B)32✓ Correct
(C)25
(D)30

Answer: (B)

Solution

For

a

: with

x^{4} \to 0

,

1 + x^{4} \approx 1 + \frac{x ^{4}}{2}

, so

1 + 1 + x^{4} \approx 2 (1 + \frac{x ^{4}}{8})

, giving

a = x \to 0 lim \frac{2 \cdot x ^{4} /8}{x ^{4}} = \frac{1}{4 2}

For

b

: with

cos x \approx 1 - \frac{x ^{2}}{2}

,

1 + cos x \approx 2 (1 - \frac{x ^{2}}{8})

, so

2 - 1 + cos x \approx \frac{2 x ^{2}}{8}

, and

sin^{2} x \approx x^{2}

:

b = x \to 0 lim \frac{x ^{2}}{2 x ^{2} /8} = 42

a b^{3} = \frac{1}{4 2} \times (42)^{3} = \frac{128 2}{4 2} = 32

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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