ExamBro
ExamBro
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:

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(6\)
Verified Solution

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:…