The value of lim x to 0 (x 2 sin 2 x/x 2 - sin 2 x ) is :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 6 April, Morning Shift — Question 37

The value of $\lim_{x \to 0} \left(\frac{x^2 \sin^2 x}{x^2 - \sin^2 x}\right)$ is :

Option A:
$2$
Option B:
$3$
Correct
Option C:
$4$
Option D:
$6$

Answer: B

Step-by-step solution

$\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{x}^{2} \sin ^{2} \mathrm{x}}{\mathrm{x}^{2}-\sin ^{2} \mathrm{x}}$ Applying expansion $\lim _{x \rightarrow 0} \frac{x^{2} \sin ^{2} x}{x^{2}-\left(x-\frac{x^{3}}{3!}+\ldots\right)^{2}}$ $\lim _{x \rightarrow 0} \frac{x^{2} \sin ^{2} x}{x^{2}-x^{2}+\frac{2 x^{4}}{3!} \ldots}=\frac{3!}{2}=3$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions