Limits, Continuity & Differentiability
L'Hôpital's Rule
Grade 12
Question:
<p>Let \(f(x)\) be a differentiable function in \([-1, \infty)\) and \(f(0) = 1\) such that \(\lim_{t \to x+1} \frac{t^2 f(x+1) - (x+1)^2 f(t)}{f(t) - f(x+1)} = 1\). Find the value of \(\lim_{x \to 1} \frac{\ln(f(x)) - \ln 2}{x-1}\).</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule to the given limit condition to extract a functional relation for f(x), then use logarithmic differentiation.
<p><strong>Step 1:</strong> The given limit $\lim_{t \to x+1} \frac{t^2 f(x+1) - (x+1)^2 f(t)}{f(t) - f(x+1)} = 1$ is in the indeterminate form $\frac{0}{0}$</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule with respect to t: $\lim_{t \to x+1} \frac{2tf(x+1) - (x+1)^2 f'(t)}{f'(t)} = 1$</p><p><strong>Step 3:</strong> This gives: $2(x+1)f(x+1) - (x+1)^2 f'(x+1) = f'(x+1)$</p><p><strong>Step 4:</strong> Rearrange: $\frac{2(x+1)f(x+1)}{f'(x+1)} = (x+1)^2 + 1$</p><p><strong>Step 5:</strong> This suggests $f(x) = \frac{x}{2}$ or analyze the structure to find $f'(x) = \frac{f(x)}{x}$</p><p><strong>Step 6:</strong> Then $\lim_{x \to 1} \frac{\ln(f(x)) - \ln 2}{x-1} = \lim_{x \to 1} \frac{f'(x)}{f(x)} = \frac{f'(1)}{f(1)} = 1$</p>
Correct Answer: 1