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

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #272

The value of

lim_{x \to 0} (\frac{x^{2} \sin^{2} x}{x^{2} - \sin^{2} x})

is:

Clue 1 of 3: Indeterminate Form

The limit is of the form $\frac{0}{0}$ as $x \to 0$ , which indicates an indeterminate form. This suggests using methods like L'Hopital's Rule or Taylor series expansions.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Indeterminate Form and Choose Method✦ Active

When $x \to 0$ , the expression becomes $\frac{0^{2} \sin^{2} 0}{0^{2} - \sin^{2} 0} = \frac{0}{0}$ , which is an indeterminate form. We can use Taylor series expansions to evaluate this limit.

Step 2: Apply Taylor Series Expansion for

\sin x

○ Expand

The Taylor series expansion for $\sin x$ around $x = 0$ is:

\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots = x - \frac{x^{3}}{6} + O (x^{5})

Now, we find the expansion for $\sin^{2} x$ :

\sin^{2} x = {(x - \frac{x^{3}}{6} + O (x^{5}))}^{2} = x^{2} {(1 - \frac{x^{2}}{6} + O (x^{4}))}^{2}

Using the binomial approximation $(1 + u)^{n} \approx 1 + n u$ for small $u$ (where $u = - \frac{x^{2}}{6}$ and $n = 2$ ):

\sin^{2} x = x^{2} (1 - 2 \cdot \frac{x^{2}}{6} + O (x^{4})) = x^{2} (1 - \frac{x^{2}}{3} + O (x^{4})) = x^{2} - \frac{x^{4}}{3} + O (x^{6})

💡 Teacher's Secret Hint

Ensure to expand to a sufficiently high order to resolve the indeterminate form. For the denominator $x^{2} - \sin^{2} x$ , we need terms up to $x^{4}$ to get a non-zero leading term.

Step 3: Substitute Expansions and Evaluate the Limit○ Expand

Substitute the expansion of $\sin^{2} x$ into the numerator and denominator of the limit expression:

Numerator: $x^{2} \sin^{2} x = x^{2} (x^{2} - \frac{x^{4}}{3} + O (x^{6})) = x^{4} - \frac{x^{6}}{3} + O (x^{8})$

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

Now, substitute these into the limit:

lim_{x \to 0} \frac{x^{4} - \frac{x^{6}}{3} + O (x^{8})}{\frac{x^{4}}{3} + O (x^{6})}

Divide both the numerator and the denominator by $x^{4}$ :

lim_{x \to 0} \frac{1 - \frac{x^{2}}{3} + O (x^{4})}{\frac{1}{3} + O (x^{2})}

As $x \to 0$ , the terms with $x^{2}$ and higher powers approach zero. Therefore, the limit is:

\frac{1}{\frac{1}{3}} = 3

💡 Teacher's Secret Hint

Remember to simplify by dividing by the lowest power of $x$ that appears in both the numerator and denominator after expansion.

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Indeterminate Form

🧠 Full Solution Path