JEE Main 2026 Apr 6 Shift 1, Mathematics Q17: Methods to Evaluate Limits

JEE Main2026Apr 6, Shift 1Mathematics

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

Solution

Use the Taylor expansion $sin x = x - \frac{x ^{3}}{6} + O (x^{5})$ , so $sin^{2} x = x^{2} - \frac{x ^{4}}{3} + O (x^{6})$ .

Numerator: $x^{2} sin^{2} x \sim x^{4}$ as $x \to 0$ .

Denominator: $x^{2} - sin^{2} x = \frac{x ^{4}}{3} + O (x^{6})$ .

$∴$ Limit $= \frac{x ^{4}}{x ^{4} /3} = 3$ .

Concept behind this question

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