Question

If \( f : \mathbb{R} \to \mathbb{R} \) is a differentiable function and \( f(3) = 6 \), then \[ \lim_{x \to 3} \frac{\displaystyle \int_{6}^{f(x)} \frac{2t \, dt}{t - 2}}{x - 3} \text{ is equal to} \]

• A. \(18f'(3)\)
• B. \(0\)
• C. \(24f'(3)\)
• D. \(3f'(3)\)

Hint:

For limits of integrals with variable upper limit, convert into derivative form directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
As \(x \to 3\), the limit takes the indeterminate form \(0/0\) since the integral from 6 to \(f(3)\) (which is 6) is zero. We apply L'Hopital's Rule along with the Leibniz Rule for differentiation of integrals.
Step 3: Detailed Explanation:
1. Let \(I = \lim_{x\to 3} \frac{\int_{6}^{f(x)} 3t dt}{x - 3}\).
2. Applying L'Hopital's Rule, we differentiate the numerator and denominator:
\[ I = \lim_{x\to 3} \frac{\frac{d}{dx} \left[ \int_{6}^{f(x)} 3t dt \right]}{\frac{d}{dx} [x - 3]} \]
3. Use the Leibniz Rule: \(\frac{d}{dx} \int_{a}^{g(x)} h(t) dt = h(g(x)) \cdot g'(x)\).
\[ \text{Numerator derivative} = [3 \cdot f(x)] \cdot f'(x) \]
\[ \text{Denominator derivative} = 1 \]
4. Evaluate the limit as \(x \to 3\):
\[ I = \frac{3 \cdot f(3) \cdot f'(3)}{1} \]
Given \(f(3) = 6\):
\[ I = 3 \times 6 \times f'(3) = 18f'(3) \]
Step 4: Final Answer:
The value of the limit is \(18f'(3)\).