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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 17

The value of lim⁡x→0(x2sin⁡2xx2−sin⁡2x)\displaystyle\lim_{x\to0}\left(\frac{x^2\sin^2x}{x^2-\sin^2x}\right) is:

Choose one option
Answer and worked solution

Official answer: B

Expansions. sin⁡2x=x2−x43+…\sin^2x=x^2-\tfrac{x^4}3+\dots, so x2−sin⁡2x≈x43x^2-\sin^2x\approx\tfrac{x^4}3. The numerator is about x4x^4.

Limit: 3.

(B)

Topics:Limits
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