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JEE Mains · Maths · STD 11 - 12. limits

જો \(f(x)\) એ સતત વિધેય છે અને  \(f\left( {\frac{9}{2}} \right) = \frac{2}{9}\), તો  \(\mathop {\lim }\limits_{x \to 0} f \left( {\frac{{1 - \cos \,3x}}{{{x^2}}}} \right)\) ની કિમંત મેળવો .

  1. A \(\frac{9}{2}\)
  2. B \(\frac{2}{9}\)
  3. C \(0\)
  4. D \(\frac{8}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{2}{9}\)

Step-by-step Solution

Detailed explanation

given that \(f\left( {\frac{9}{2}} \right) = \frac{2}{9}\) \(\,\mathop {\lim }\limits_{x \to 0} f\left( {\frac{{1 - \cos 3x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{1 - \cos 3x}}} \right)\)…
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