JEE Mains · Maths · STD 11 - 12. limits
If \(f(x)\) is continuous and \(f\left( {\frac{9}{2}} \right) = \frac{2}{9}\), then \(\mathop {\lim }\limits_{x \to 0} f \left( {\frac{{1 - \cos \,3x}}{{{x^2}}}} \right)\) is equal to:
- A \(\frac{9}{2}\)
- B \(\frac{2}{9}\)
- C \(0\)
- D \(\frac{8}{9}\)
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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