Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$\lim _{x \rightarrow 1^{+}} \frac{\int_{1}^{x} \sin t d t}{x-1}$$

Step-by-Step Solution

Answer

The limit is \( \sin 1 \).

Step 1: Understand the Problem

We need to find the limit of the function \( \frac{\int_{1}^{x} \sin t \, dt}{x-1} \) as \( x \to 1^+ \). First, let's consider the form of this expression when \( x \to 1^+ \).

Step 2: Evaluate the Numerator and Denominator at the Point of Interest

As \( x \to 1^+ \), the denominator \( x-1 \to 0 \). Similarly, the numerator \( \int_{1}^{x} \sin t \, dt \to 0 \) because it is the integral from 1 to 1. This gives an indeterminate form \( \frac{0}{0} \).

Step 3: Verify Indeterminate Form

Having confirmed the \( \frac{0}{0} \) form, we can apply l'Hôpital's Rule. This involves differentiating the numerator and denominator separately and then finding the limit of the new expression.

Step 4: Differentiate the Numerator and Denominator

The derivative of the numerator with respect to \( x \) is \( \sin x \) (by the Fundamental Theorem of Calculus). The derivative of the denominator is \( 1 \). Hence, the limit becomes \( \lim _{x \rightarrow 1^{+}} \frac{\sin x}{1} \).

Step 5: Apply the Limit

Now find \( \lim _{x \rightarrow 1^{+}} \sin x \), which equals \( \sin 1 \), since \( 1 \) is within the domain of the sine function. Therefore, the limit is \( \sin 1 \).

Key Concepts

Indeterminate Forms

When dealing with limits, it's crucial to identify whether you're working with an indeterminate form. These forms occur when substituting the limit's value into a function leads to an unclear or undefined expression. One common indeterminate form is \(\frac{0}{0}\).

In the given exercise, we encountered such a scenario. As \(x\) approaches \(1^+\), both the numerator \(\int_{1}^{x} \sin t \, dt\) and the denominator \(x - 1\) trend towards zero. This results in the \(\frac{0}{0}\) form, giving no direct answer to the limit question.

The presence of indeterminate forms is why techniques like l'Hôpital's Rule are so valuable. By identifying these forms, we can justify using such methods to find clearer answers by further evaluating the function through derivatives.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, providing a powerful tool for evaluating integrals. It states that if a function is continuous over an interval, then the function's indefinite integral is differential and its derivative yields the original function.

In this exercise, we applied the theorem to find the derivative of the numerator \(\int_{1}^{x} \sin t \, dt\). According to the theorem, differentiating the integral \(\int_{1}^{x} \sin t \, dt\) with respect to \(x\) results in \(\sin x\).

This step is vital as it transformed our bottleneck expression into something manageable for l'Hôpital's Rule, allowing us to substitute it back into the function for a clearer evaluation of the limit.

Calculus Limits

Limits in calculus form the foundation of understanding functions at boundaries or particular points. When dealing with calculus limits, the goal is to find the function's value as the variable approaches a specific number or infinity.

In our example, the focus was on evaluating the limit as \(x \to 1^+\). Initially, both parts of the fraction yielded zero, an indeterminate form. After applying l'Hôpital's Rule, we simplified the problem to analyzing \(\lim _{x \rightarrow 1^{+}} \sin x\).

This simplified expression is direct as \(x\) approaches \(1^+\), making the result easier to discern. Calculating limits, particularly in situations requiring simplification or transformation, is a key skill in calculus. Recognizing when to apply rules, like l'Hôpital's Rule, ensures accurate evaluations and strengthens problem-solving abilities.