Let `f : R to R ` be a continuously differentiable function such that `f(2) = 6` and `f'(2) = 1/48`. If `int_(6)^(f(x)) 4t^(3) dt = (x-2) g(x)` than `lim_( x to 2) g(x)` is equal to

Options

A. 12
B. 18
C. 36
D. 24

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Understand the given information
We know that:
- \( f(2) = 6 \)
- \( f'(2) = \frac{1}{48} \)
- The integral condition is given as:
\[ \int_{6}^{f(x)} 4t^{3} \, dt = (x - 2) g(x) \]

### Step 2: Express \( g(x) \)
From the integral condition, we can express \( g(x) \) as:
\[ g(x) = \frac{1}{x - 2} \int_{6}^{f(x)} 4t^{3} \, dt \]

### Step 3: Evaluate the limit
We need to find:
\[ \lim_{x \to 2} g(x) \]
Substituting \( x = 2 \) directly into \( g(x) \) gives us an indeterminate form \( \frac{0}{0} \) because:
- \( f(2) = 6 \) (upper limit equals lower limit)
- \( x - 2 = 0 \)

### Step 4: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator with respect to \( x \):
\[ \lim_{x \to 2} g(x) = \lim_{x \to 2} \frac{\int_{6}^{f(x)} 4t^{3} \, dt}{x - 2} \]

Using Leibniz's rule to differentiate the integral:
\[ \frac{d}{dx} \left( \int_{6}^{f(x)} 4t^{3} \, dt \right) = 4(f(x))^{3} f'(x) \]
Thus, we have:
\[ \lim_{x \to 2} g(x) = \lim_{x \to 2} \frac{4(f(x))^{3} f'(x)}{1} \]

### Step 5: Substitute the values
Now substituting \( x = 2 \):
- \( f(2) = 6 \)
- \( f'(2) = \frac{1}{48} \)

So we get:
\[ = 4 \cdot (6)^{3} \cdot \frac{1}{48} \]

### Step 6: Calculate the value
Calculating \( (6)^{3} = 216 \):
\[ = 4 \cdot 216 \cdot \frac{1}{48} \]
\[ = \frac{864}{48} = 18 \]

### Conclusion
Thus, the required limit is:
\[ \lim_{x \to 2} g(x) = 18 \]