LimitsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Limit x²sin²x/(x²-sin²x) = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The value of lim⁡x→0(x2sin⁡2xx2−sin⁡2x)\displaystyle\lim_{x\to0}\left(\dfrac{x^2\sin^2 x}{x^2-\sin^2 x}\right) is
A22
B33correct
C44
D66
Solution
Step 1: sin⁡x=x−x36+⋯\sin x=x-\dfrac{x^3}{6}+\cdots
sin⁡2x=(x−x36+⋯ )2=x2−x43+⋯ .\sin^2 x=\left(x-\dfrac{x^3}{6}+\cdots\right)^2=x^2-\dfrac{x^4}{3}+\cdots.
 Step 2: x2−sin⁡2x=x2−(x2−x43+⋯ )=x43+⋯x^2-\sin^2 x=x^2-\left(x^2-\dfrac{x^4}{3}+\cdots\right)=\dfrac{x^4}{3}+\cdots Step 3: x2sin⁡2x=x2(x2−x43+⋯ )=x4+⋯x^2\sin^2 x=x^2\left(x^2-\dfrac{x^4}{3}+\cdots\right)=x^4+\cdots Step 4:
∴lim⁡x→0x2sin⁡2xx2−sin⁡2x=lim⁡x→0x4+⋯x43+⋯=11/3=3.\therefore\lim_{x\to0}\dfrac{x^2\sin^2 x}{x^2-\sin^2 x}=\lim_{x\to0}\dfrac{x^4+\cdots}{\tfrac{x^4}{3}+\cdots}=\dfrac{1}{1/3}=3.
 Correct answer: (2)
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