Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x} \sqrt{t} \cos t d t}{x^{2}} $$

Step-by-Step Solution

Answer

The limit approaches infinity as \(x\) approaches \(0^{+}\).

Step 1: Identify the Form of the Limit

The given expression is a limit: \[ \lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x} \sqrt{t} \cos t \, dt}{x^{2}} \] It suggests the use of L'Hôpital's Rule, since it is in the indeterminate form \(\frac{0}{0}\) as \(x\) approaches \(0^{+}\).

Step 2: Differentiate the Numerator

First, we need to differentiate the numerator, which is an integral: \[ F(x) = \int_{0}^{x} \sqrt{t} \cos t \, dt \]By the Fundamental Theorem of Calculus, we have:\[ F'(x) = \sqrt{x} \cos x \]

Step 3: Differentiate the Denominator

Differentiate the denominator, \(x^2\), with respect to \(x\):\[ \frac{d}{dx}\left(x^2\right) = 2x \]

Step 4: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule, which can be used for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) forms:\[\lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x} \sqrt{t} \cos t \, dt}{x^{2}} = \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x} \cos x}{2x} \]

Step 5: Simplify the Expression

Simplify the expression from L'Hôpital's Rule:\[ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x} \cos x}{2x} = \lim _{x \rightarrow 0^{+}} \frac{\cos x}{2 \sqrt{x}} \]

Step 6: Evaluate the Simplified Limit

As \(x\) approaches \(0^{+}\), \(\cos x\) approaches \(1\), resulting in:\[ \lim _{x \rightarrow 0^{+}} \frac{1}{2 \sqrt{x}} \]This expression approaches infinity as \(x\) approaches \(0^{+}\) from the right since \(\sqrt{x}\) approaches zero.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to resolve limits that initially present indeterminate forms. When a limit results in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule can assist in simplifying and computing these limits. It states that if \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \), or \( \lim_{x \to c} f(x) = \infty \) and \( \lim_{x \to c} g(x) = \infty \), then:

• \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \)

provided the limit on the right-hand side exists.
In the given exercise, the initial form of \( \frac{\int_{0}^{x} \sqrt{t} \cos t \, dt}{x^{2}} \) results in \( \frac{0}{0} \). By applying L'Hôpital's Rule, we differentiated the numerator and denominator individually to simplify the expression into a solvable format. This transformed the limit into \( \lim_{x \to 0^+} \frac{\sqrt{x} \cos x}{2x} \), allowing further assessments of the function behavior as \( x \) approaches zero.
L'Hôpital's Rule is a strategic approach when standard algebraic methods fail with indeterminate forms.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two fundamental operations in calculus. It states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \( [a, b] \), then:

• \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \)

This implies that differentiation and integration are inverses of each other.
In the exercise, the problem involves the computation of the derivative of an integral. We use the Fundamental Theorem of Calculus, which tells us that if \( F(x) = \int_{0}^{x} \sqrt{t} \cos t \, dt \), then \( F'(x) = \sqrt{x} \cos x \).
Knowing this property allows us to focus on differentiating the integral as a function of \( x \). This step is crucial in applying L'Hôpital's Rule since differentiating the integral gives us the numerator required to simplify the indeterminate limit.
Without understanding this theorem, solving the limit would be much more complex. It simplifies the process by providing a straightforward approach to differentiating integrals.

Indeterminate Forms

Indeterminate forms occur when evaluating a limit results in an unclear or undefined expression. Common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \). Such forms signify that the expression does not straightforwardly point to a clear result.
In the limit \( \lim_{x \to 0^+} \frac{\int_{0}^{x} \sqrt{t} \cos t \, dt}{x^2} \), both the numerator and denominator approach zero as \( x \to 0^+ \). This creates a \( \frac{0}{0} \) indeterminate form, which is a signal that direct evaluation isn't possible without further computation.

• Strategies for resolving indeterminate forms include algebraic manipulation and applying calculus techniques like L'Hôpital's Rule.

These approaches help transform the expressions into a form where limits can be easily evaluated. Understanding indeterminate forms is essential in recognizing when and how to apply the correct mathematical techniques to derive a meaningful result.
It is a fundamental part of calculus that guides the process of evaluating complex limits.