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Calculus

Limits, Standard Limits

jee_main_2026_april_6_shift_1

Grade None

Question:

Find the value of lim_{x→0} (sin 5x - sin 3x)/(tan 2x).

A. 1

B. 2

C. 3

D. 4

Step-by-Step Solution

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

Step 1: sin 5x - sin 3x = 2 cos 4x sin x. Step 2: Expression = 2 cos 4x sin x / tan 2x. Step 3: As x→0, sin x/tan 2x ≈ x/(2x) = 1/2. Step 4: Limit = 2 cos 0 × 1/2 = 2 × 1 × 1/2 = 1.

Correct Answer: A

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