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Calculus

Limits, Standard Limits

jee_main_2026_april_6_shift_2

Grade None

Question:

Find the value of lim_{x→0} (sin 4x - sin 2x)/(tan 3x).

A. 1/3

B. 2/3

C. 1

D. 4/3

Step-by-Step Solution

Key Concept: Use sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2).

Step 1: sin 4x - sin 2x = 2 cos 3x sin x. Step 2: Expression = 2 cos 3x sin x / tan 3x. Step 3: As x→0, sin x/tan 3x ≈ x/(3x) = 1/3. Step 4: Limit = 2 cos 0 × 1/3 = 2 × 1 × 1/3 = 2/3.

Correct Answer: B

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