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Step-by-Step Solution
Step 1: Express the limit
We want to evaluate
$$
\lim_{x \to 2} \frac{\int_{6}^{f(x)} 2t \, dt}{x - 2}.
$$
Since \(f(2) = 6\), notice that as \(x \to 2\), the numerator becomes
\(\int_{6}^{6} 2t\,dt = 0\) and the denominator also goes to \(0\). This creates a \(\frac{0}{0}\) indeterminate form.
Step 2: Recognize the indeterminate form and decide on a method
The \(\frac{0}{0}\) form suggests using L'Hôpital's Rule, which requires us to differentiate the numerator and denominator with respect to \(x\).
Step 3: Differentiate the numerator
The numerator is
\(\int_{6}^{f(x)} 2t \, dt.\)
Using the Leibniz rule (differentiation under the integral with variable upper limit), we get:
$$
\frac{d}{dx} \left[ \int_{6}^{f(x)} 2t \, dt \right]
= 2f(x) \cdot \frac{d}{dx}[\,f(x)\,]
= 2f(x) \, f'(x).
$$
Step 4: Differentiate the denominator
The denominator is \(x - 2\). Its derivative with respect to \(x\) is \(1\).
Step 5: Apply L'Hôpital's Rule
By L'Hôpital's Rule,
$$
\lim_{x \to 2} \frac{\int_{6}^{f(x)} 2t \, dt}{x - 2}
= \lim_{x \to 2} \frac{2f(x) \, f'(x)}{1}
= 2f(x)\, f'(x)\Big\vert_{x=2}.
$$
Step 6: Evaluate at \(x = 2\)
Since \(f(2) = 6,\) substitute \(x = 2\) into the expression:
$$
2 \cdot f(2) \cdot f'(2) = 2 \cdot 6 \cdot f'(2) = 12f'(2).
$$
Hence, the limit is \(12f'(2).\)
Step 7: Conclude the correct answer
The value of the given limit is
\(\boxed{12f'(2)}.\)
So, the correct choice is \(12f'(2).\)
