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 given expression involves the limit of an integral. We are tasked with finding the value of this limit.

Step 2: The integral is given as:

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

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

Step 3: First, rewrite the integral as follows:

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

Notice that as \( x \to 1 \), \( f(x) \to 4 \). Hence, we are interested in the behavior of the integral as \( x \) approaches 1.

Step 4: The integrand has the form \( \frac{2t}{x-1} \), which suggests that the integral evaluates to a result proportional to \( \frac{1}{x-1} \).

Step 5: Differentiating the integral expression with respect to \( x \), we get:

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

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

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

Thus, the value of the given expression is \( \boxed{16} \).

Answer 2

Evaluating the Limit $\lim_{x \to 1} \frac{1}{x-1} \int_{4}^{f(x)} 2t \, dt$

We are given a differentiable function $f: \mathbb{R} \to \mathbb{R}$ with $f(1) = 4$ and $f'(1) = 2$. We need to find the value of the limit:

$$ L = \lim_{x \to 1} \frac{1}{x-1} \int_{4}^{f(x)} 2t \, dt $$

Step 1: Evaluate the definite integral

First, evaluate the integral with respect to $t$:

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

Step 2: Substitute the result back into the limit expression

Now, substitute this back into the limit:

$$ L = \lim_{x \to 1} \frac{(f(x))^2 - 16}{x-1} $$

Step 3: Check for indeterminate form

As $x \to 1$, the numerator approaches $(f(1))^2 - 16 = 4^2 - 16 = 16 - 16 = 0$.

As $x \to 1$, the denominator approaches $1 - 1 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we can use L'Hôpital's Rule.

Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator with respect to $x$:

Derivative of the numerator: $\frac{d}{dx}((f(x))^2 - 16) = 2f(x)f'(x)$

Derivative of the denominator: $\frac{d}{dx}(x-1) = 1$

Now, apply the limit to the ratio of the derivatives:

$$ L = \lim_{x \to 1} \frac{2f(x)f'(x)}{1} $$

Step 5: Substitute the given values

Substitute the given values $f(1) = 4$ and $f'(1) = 2$:

$$ L = 2f(1)f'(1) = 2(4)(2) = 16 $$

Therefore, the value of the limit is 16.

Final Answer: (A) 16