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JEE Mains · Maths · STD 11 - 12. limits

\(\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\,\sin \,x}}{{\sqrt {{x^2} + 2\sin \,x + 1}  - \sqrt {{{\sin }^2}\,x - x + 1} }}\) is

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

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}\) \( = \mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{{x^2} + 2\sin x + 1 - {{\sin }^2}x - x + 1}}\)…
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