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

Options

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

Answer

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

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

### Step 1: Identify the form of the limit
As \( x \to 1 \), the upper limit \( f(x) \) approaches \( f(1) = 4 \). Therefore, both the upper limit and the denominator \( x - 1 \) approach 0, leading to a \( \frac{0}{0} \) indeterminate form.

**Hint:** Check the behavior of the upper limit and the denominator as \( x \) approaches 1.

### Step 2: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule. This requires us to differentiate the numerator and the denominator.

**Hint:** Remember that L'Hôpital's Rule states that if you have a limit of the form \( \frac{0}{0} \), you can take the derivative of the numerator and the denominator.

### Step 3: Differentiate the numerator
Using the Fundamental Theorem of Calculus, the derivative of the integral with respect to \( x \) is:

\[ \frac{d}{dx} \left( \int_{4}^{f(x)} \frac{2t}{x-1} \, dt \right) = \frac{2f(x)}{x-1} \cdot f'(x) - 0 \]

Here, \( \frac{2t}{x-1} \) is evaluated at the upper limit \( f(x) \) and multiplied by the derivative of the upper limit \( f'(x) \).

**Hint:** When differentiating an integral with variable limits, remember to multiply by the derivative of the upper limit.

### Step 4: Differentiate the denominator
The derivative of the denominator \( x - 1 \) is simply:

\[ \frac{d}{dx}(x - 1) = 1 \]

**Hint:** The derivative of a linear function is constant.

### Step 5: Rewrite the limit
Now we can rewrite the limit using L'Hôpital's Rule:

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

### Step 6: Evaluate the limit
Substituting \( x = 1 \):

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

**Hint:** Substitute the known values directly into the expression after applying L'Hôpital's Rule.

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

\[ \boxed{8f'(1)} \]