JEE Mains · Maths · STD 11 - 12. limits
The value of \(\lim_{x \to 0}\left(\dfrac{x^2\sin^2 x}{x^2 - \sin^2 x}\right)\) is:
- A \(2\)
- B \(3\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
Let \(L = \lim_{x \to 0} \dfrac{x^2\sin^2 x}{x^2 - \sin^2 x}\) Dividing the numerator and the denominator by \(x^4\), we get: \(L = \lim_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\dfrac{x^2 - \sin^2 x}{x^4}}\) Factorizing the denominator:…
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