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Calculus

Limits, Standard Limits

jee_main_2026_jan_22_shift_1

Grade None

Question:

Find the value of \(\lim_{x\to 0} \frac{\sin 4x - \sin 2x}{\tan 3x}\).

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 = 2/3.

Correct Answer: B

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