Q.17. The value of is:

Sol. $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$ Applying expansion $\sin x = x - \frac{x^3}{3!} + O(x^5)$. So, $\sin^2 x = (x - \frac{x^3}{3!} + ...)^2 = x^2 - \frac{2x^4}{3!} + O(x^6)$. Then, $\lim_{x \to 0} \frac{x^2 (x^2 - \frac{2x^4}{3!} + ...)}{x^2 - (x^2 - \frac{2x^4}{3!} + ...)} = \lim_{x \to 0} \frac{x^4 - \frac{2x^6}{3!} + ...}{\frac{2x^4}{3!} - O(x^6)} = \frac{1}{\frac{2}{3!}} = \frac{3!}{2} = \frac{6}{2} = 3$.

What is the correct answer for JEE Main Mathematics Q17?

The correct answer is B. Sol. $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$ Applying expansion $\sin x = x - \frac{x^3}{3!} + O(x^5)$. So, $\sin^2 x = (x - \frac{x^3}{3!} + ...)^2 = x^2 - \frac{2x^4}{3!} + O(x^6)$. Then, $\lim_{x \to 0} \frac{x^2 (x^2 - \frac{2x^4}{3!} + ...)}{x^2 - (x^2 - \frac{2x^4}{3!} + ...)} = \lim_{x \to 0} \frac{x^4 - \frac{2x^6}{3!} + ...}{\frac{2x^4}{3!} - O(x^6)} = \frac{1}{\frac{2}{3!}}

Q.17. The value of is:

The correct answer is B. Sol. $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$ Applying expansion $\sin x = x - \frac{x^3}{3!} + O(x^5)$. So, $\sin^2 x = (x - \frac{x^3}{3!} + ...)^2 = x^2 - \frac{2x^4}{3!} + O(x^6)$. Then, $\lim_{x \to 0} \frac{x^2 (x^2 - \frac{2x^4}{3!} + ...)}{x^2 - (x^2 - \frac{2x^4}{3!} + ...)} = \lim_{x \to 0} \frac{x^4 - \frac{2x^6}{3!} + ...}{\frac{2x^4}{3!} - O(x^6)} = \frac{1}{\frac{2}{3!}}

JEE Main 2026… PYQ · With Answer · Free

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Q.17. The value of $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$ is:

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