Let `f : R to R` be a differentiable function and `f(1) = 4`. Then, the value of `lim_(x to 1)int_(4)^(f(x))(2t)/(x-1)dt` is :

Options

A. `8 f'(1)`
B. `4 f'(1)`
C. `2 f'(1)`
D. `f'(1)`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 1} \int_{4}^{f(x)} \frac{2t}{x-1} dt \]

Given that \( f(1) = 4 \), we can start by rewriting the integral.

### Step 1: Rewrite the Integral
The integral can be rewritten as:

\[ \lim_{x \to 1} \frac{1}{x-1} \int_{4}^{f(x)} 2t \, dt \]

### Step 2: Evaluate the Integral
Now, we need to evaluate the integral \( \int_{4}^{f(x)} 2t \, dt \):

\[ \int 2t \, dt = t^2 \]

Thus,

\[ \int_{4}^{f(x)} 2t \, dt = [t^2]_{4}^{f(x)} = f(x)^2 - 4^2 = f(x)^2 - 16 \]

### Step 3: Substitute Back into the Limit
Substituting this back into our limit gives:

\[ \lim_{x \to 1} \frac{f(x)^2 - 16}{x - 1} \]

### Step 4: Apply L'Hôpital's Rule
As \( x \to 1 \), both the numerator and denominator approach 0, resulting in a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

- The derivative of the numerator \( f(x)^2 - 16 \) is \( 2f(x)f'(x) \) (using the chain rule).
- The derivative of the denominator \( x - 1 \) is \( 1 \).

Thus, we have:

\[ \lim_{x \to 1} \frac{2f(x)f'(x)}{1} \]

### Step 5: Evaluate the Limit
Now, substituting \( x = 1 \):

\[ \lim_{x \to 1} 2f(x)f'(x) = 2f(1)f'(1) \]

Since \( f(1) = 4 \):

\[ = 2 \cdot 4 \cdot f'(1) = 8f'(1) \]

### Final Answer
Therefore, the value of the limit is:

\[ 8f'(1) \]