Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\lim _{x \rightarrow 1} \frac{1}{x-1} \int_{1}^{x} \frac{1+t}{2+t} d t\).

Step-by-Step Solution

Answer

The limit is \( \frac{2}{3} \).

Step 1: Understand the Problem

We need to calculate the limit \( \lim _{x \rightarrow 1} \frac{1}{x-1} \int_{1}^{x} \frac{1+t}{2+t} \, dt \). This expression involves both a limit and a definite integral, which suggests we might be able to use l'Hôpital's Rule.

Step 2: Recognize the Indeterminate Form

First, recognize that as \( x \rightarrow 1 \), both the numerator \( \int_{1}^{x} \frac{1+t}{2+t} \, dt \) and the denominator \( x - 1 \) approach zero, creating a \( \frac{0}{0} \) indeterminate form. This is suitable for applying l'Hôpital's Rule.

Step 3: Differentiate the Numerator and Denominator

By the Fundamental Theorem of Calculus, the derivative of the numerator \( \int_{1}^{x} \frac{1+t}{2+t} \, dt \) with respect to \( x \) is simply \( \frac{1+x}{2+x} \). The derivative of the denominator, \( x-1 \), with respect to \( x \) is 1.

Step 4: Apply l'Hôpital's Rule

Substitute the derivatives into l'Hôpital's Rule: \[ \lim_{x \rightarrow 1} \frac{\frac{1+x}{2+x}}{1} = \lim_{x \rightarrow 1} \frac{1+x}{2+x} \]

Step 5: Evaluate the Limit

Evaluate the limit expression. Plug \( x = 1 \) into \( \frac{1+x}{2+x} \) to get:\[ \lim_{x \rightarrow 1} \frac{1+x}{2+x} = \frac{1+1}{2+1} = \frac{2}{3} \]

Step 6: Conclusion: State the Final Result

The value of the limit is \( \frac{2}{3} \). This concludes the process of solving the problem.

Key Concepts

l'Hôpital's Rule

When faced with limits that produce an indeterminate form like \( \frac{0}{0} \), l'Hôpital's Rule becomes a powerful tool. It allows you to differentiate the numerator and the denominator separately to find the limit of the function.

• First, check that both the numerator and the denominator approach zero or infinity as \( x \) approaches a specified value.
• Once confirmed, differentiate the numerator and the denominator with respect to \( x \).
• Re-evaluate the limit using these new derivatives. If the result is no longer indeterminate, you can find a finite limit.

In our exercise, both the integral and \( x-1 \) tend to zero as \( x \to 1 \), meeting the criteria for l'Hôpital's Rule. By differentiating, we transform a complex problem into a straightforward evaluation, simplifying the computation of limits in calculus problems.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration. It tells us that the derivative of an integral function essentially returns us to the original function.

• Part 1 states that if \( F \) is an antiderivative of \( f \) over an interval, then the integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).
• Part 2, which is often used in these contexts, shows that the derivative of the integral of a function is the function itself, \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \).

In the given exercise, the Fundamental Theorem of Calculus allows us to take the derivative of the integral \( \int_{1}^{x} \frac{1+t}{2+t} \, dt \) with ease, resulting in the function \( \frac{1+x}{2+x} \). This simplifies the application of l'Hôpital's Rule.

Definite Integrals

Definite integrals play a crucial role in calculating the area under a curve between two points. They are not only about computing areas but also have applications in other fields of mathematics and science.

• The definite integral from \( a \) to \( b \) of a function \( f(t) \) is denoted as \( \int_{a}^{b} f(t) \, dt \).
• It represents the net area between the x-axis and the curve \( f(t) \), taking into account regions below the x-axis as negative areas.

In our context, the definite integral \( \int_{1}^{x} \frac{1+t}{2+t} \, dt \) helps form the numerator of the ratio. This allows us to apply calculus techniques to explore the limit behavior of the expression as \( x \to 1 \), making integrals a versatile and critical part of calculus.