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 limit problem, we need to evaluate the following expression:

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

### Step 1: Recognize the form of the limit
As \( x \to 1 \), both the numerator and denominator approach 0, which gives us an indeterminate form \( \frac{0}{0} \). This suggests that we can apply L'Hôpital's Rule.

**Hint:** Check if the limit results in an indeterminate form to decide if L'Hôpital's Rule can be applied.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

\[ \lim_{x \to 1} \frac{d}{dx} \left( \int_{4}^{f(x)} \frac{2t}{x-1} dt \right) \bigg/ \frac{d}{dx}(x-1) \]

### Step 3: Differentiate the numerator
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate the numerator:

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

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

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

### Step 5: Rewrite the limit
Now we can rewrite the limit using the derivatives we found:

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

### Step 6: Evaluate the limit
Now we substitute \( x = 1 \) into the expression:

\[ = 2f(1) f'(1) \]

Given that \( f(1) = 4 \), we have:

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

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

\[ 8 f'(1) \]

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