<p>Let \(f: R \to R\) be a differentiable function satisfying \(f'(3) + f'(2) = 0\). Then \(\lim_{x \to 0} \left(\dfrac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)}\right)^{\frac{1}{x}}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator and denominator are of the form (1 + derivative terms), allowing you to use the exponential limit definition: lim(1 + u)^(1/x) = e^(lim u/x). The constraint f'(3) + f'(2) = 0 ensures the exponent doesn't vanish.
<p><strong>Step 1:</strong> As x→0, the numerator approaches 1 + f'(3)·1 = 1 + f'(3) and denominator approaches 1 + f'(2)·1 = 1 + f'(2), giving form 1^∞.</p><p><strong>Step 2:</strong> Rewrite the limit as e^(lim(x→0) [ln(1+f(3+x)-f(3)) - ln(1+f(2-x)-f(2))]/x)</p><p><strong>Step 3:</strong> For small arguments, ln(1+u) ≈ u, so:<br/>Exponent = lim(x→0) [f(3+x)-f(3) - (f(2-x)-f(2))]/x</p><p><strong>Step 4:</strong> Split the limit:<br/>= lim(x→0) [f(3+x)-f(3)]/x - lim(x→0) [f(2-x)-f(2)]/x<br/>= f'(3) - (-f'(2)) [note: lim(x→0) [f(2-x)-f(2)]/x = -f'(2)]<br/>= f'(3) + f'(2) = 0</p><p><strong>Step 5:</strong> Therefore the answer is e^0 = <strong>1</strong></p><p>∴ Answer: A</p>
Correct Answer: A