Question

Let \(f : \mathbb{R} \to \mathbb{R}\) be a differentiable function and \(f(1) = 4\). Then the value of
\[ \lim_{x \to 1} \int_{4}^{f(x)} \frac{2t}{x - 1} \, dt \]

• A. 16
• B. 8
• C. 4
• D. 2

Hint:

When dealing with limits of integrals, consider the use of the Fundamental Theorem of Calculus and differentiate under the integral sign if necessary.

Answer 1

The Correct Option is A

Step 1: The problem requires finding the limit of an integral expression.

Step 2: The integral is:

\[ \int_4^{f(x)} \frac{2t}{x-1} \, dt \]

Given \( f(x) \) is differentiable and \( f(1) = 4 \), we can use the Fundamental Theorem of Calculus.

Step 3: The integral remains:

\[ \int_4^{f(x)} \frac{2t}{x-1} \, dt \]

As \( x \to 1 \), \( f(x) \to 4 \). We analyze the integral's behavior as \( x \) approaches 1.

Step 4: The integrand \( \frac{2t}{x-1} \) suggests the integral is proportional to \( \frac{1}{x-1} \).

Step 5: Differentiating the integral with respect to \( x \):

\[ \frac{d}{dx} \left( \int_4^{f(x)} \frac{2t}{x-1} \, dt \right) = \frac{2f(x)}{x-1} \cdot f'(x) \]

Step 6: Substituting \( f(1) = 4 \) and \( f'(1) = 2 \), evaluate the limit at \( x = 1 \):

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

Therefore, the expression's value is \( \boxed{16} \).