JEE Mains · Maths · STD 11 - 12. limits
Let \(f : R \to R\) be a differentiable function satisfying \(f’’(3) + f’(2) = 0\). Then \(\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}\) is equal to
- A \(e^2\)
- B \(1\)
- C \(e\)
- D \(e^{-1}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\({1^\infty }\) form \(k = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{f\left( {3 + x} \right) - f\left( {2x} \right) - f\left( 3 \right)\left( {f\left( 2 \right)} \right)}}{{x\left( {1 + f\left( {2 - x} \right) - f\left( 2 \right)} \right)}}} \right)\)…
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