Problem 34 - Calculus Early Transcendentals

Question

Use the Squeeze Theorem to show that $$\lim _{x \rightarrow 0} \sqrt{x^{3}+x^{2}} \sin \frac{\pi}{x}=0$$ Illustrate by graphing the functions \(f, g,\) and \(h\) (in the notation of the Squeeze Theorem) on the same screen.

Step-by-Step Solution

Answer

The limit is 0, verified by the Squeeze Theorem.

Step 1: Understand the Squeeze Theorem

The Squeeze Theorem states that if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in some interval containing \(c\) (except possibly at \(c\)) and \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\). In this problem, we need to define functions \(f(x)\), \(g(x)\), and \(h(x)\) to apply this theorem.

Step 2: Define the Functions for Squeeze Theorem

Define \(g(x) = \sqrt{x^3 + x^2} \sin \frac{\pi}{x}\). We need to find functions \(f(x)\) and \(h(x)\) such that \(f(x) \leq g(x) \leq h(x)\). Since \(\sin \frac{\pi}{x}\) oscillates between -1 and 1, we have: \( -\sqrt{x^3 + x^2} \leq \sqrt{x^3 + x^2} \sin \frac{\pi}{x} \leq \sqrt{x^3 + x^2} \)

Step 3: Apply the Limit to the Boundary Functions

Now, we need to find the limits of \(f(x) = -\sqrt{x^3 + x^2}\) and \(h(x) = \sqrt{x^3 + x^2}\) as \(x\) approaches 0. Calculate \(\lim_{x \to 0} -\sqrt{x^3 + x^2} = \lim_{x \to 0} \sqrt{x^3 + x^2} = 0\) because \(x^3 + x^2\) becomes very small as \(x\) approaches 0.

Step 4: Conclude Using the Squeeze Theorem

Since both boundary functions \(f(x) = -\sqrt{x^3 + x^2}\) and \(h(x) = \sqrt{x^3 + x^2}\) approach 0 as \(x\) approaches 0, and since \(f(x) \leq g(x) \leq h(x)\), by the Squeeze Theorem \(\lim_{x \to 0} \sqrt{x^3 + x^2} \sin \frac{\pi}{x} = 0\).

Step 5: Graph the Functions for Visualization

For better understanding, graph the functions \(f(x) = -\sqrt{x^3 + x^2}\), \(g(x) = \sqrt{x^3 + x^2} \sin \frac{\pi}{x}\), and \(h(x) = \sqrt{x^3 + x^2}\). Notice that the graph of \(g(x)\) is squeezed between \(f(x)\) and \(h(x)\) as \(x\) approaches 0, illustrating the Squeeze Theorem's effect.

Key Concepts

Limits

A limit represents the value a function approaches as the input approaches a certain point. In calculus, understanding limits is crucial for defining derivatives and integrals. This problem uses the Squeeze Theorem, which relies heavily on the concept of limits.

In the given exercise, we aim to evaluate \[ \lim _{x \rightarrow 0} \sqrt{x^{3}+x^{2}} \sin \frac{\pi}{x} = 0 \]. This requires us to find the behavior of this function as x approaches 0. The Squeeze Theorem helps us achieve this by comparing the function with two simpler functions whose limits are known.

• Understanding the Boundary: We note that the expression inside the limit involves a product: \(\sqrt{x^3 + x^2}\) and \(\sin \frac{\pi}{x}\). As \(x\) approaches 0, \(\sqrt{x^3 + x^2}\) tends to 0 while \(\sin \frac{\pi}{x}\) remains bounded between -1 and 1.
• Application of Limits: By applying limits to the boundary functions, we establish that their values both approach 0 as \(x\) approaches 0.

Understanding limits provide the backbone for using the Squeeze Theorem, allowing us to precisely determine the behavior of more complex functions.

Boundary Functions

Boundary functions are essential to leverage the Squeeze Theorem effectively. These functions are constructed to "sandwich" the target function, providing upper and lower bounds, ensuring that as they converge to a common limit, so does the target function.

For this problem, we define:

• \(f(x) = -\sqrt{x^3 + x^2}\): This function provides a lower boundary, exploiting the fact that \(\sin \frac{\pi}{x}\) oscillates between -1 and 1.
• \(h(x) = \sqrt{x^3 + x^2}\): This function creates an upper boundary, completing the "squeeze."

Both \(f(x)\) and \(h(x)\) are chosen because they bound \(g(x) = \sqrt{x^3 + x^2} \sin \frac{\pi}{x}\), making it possible to apply the Squeeze Theorem.

By computing the limits of these boundary functions as \(x\) approaches 0, we observe:

• \(\lim_{x \to 0} f(x) = \lim_{x \to 0} -\sqrt{x^3 + x^2} = 0\)
• \(\lim_{x \to 0} h(x) = \lim_{x \to 0} \sqrt{x^3 + x^2} = 0\)

Therefore, the boundary functions ensure that \(g(x)\) is also squeezed to the limit 0.

Oscillating Functions

Oscillating functions, like \(\sin \frac{\pi}{x}\) in our example, are crucial to understanding this calculus problem. An oscillating function is one whose values continually vacillate between certain bounds as the input changes.

In this exercise, \(\sin \frac{\pi}{x}\) does not settle on a single value as \(x\) approaches 0. Instead, it oscillates indefinitely between -1 and 1. This oscillation might make it seem challenging to determine the limit of the entire function \(\sqrt{x^{3}+x^{2}} \sin \frac{\pi}{x}\).

However, the Squeeze Theorem helps manage this behavior. By establishing upper and lower boundary functions that both converge to 0, the oscillating nature of \(\sin \frac{\pi}{x}\) becomes irrelevant to the limit. The non-oscillating factor \(\sqrt{x^3 + x^2}\) ensures the whole expression is squeezed to 0.

• Despite the oscillation, the amplitude is modulated by a factor that vanishes as \(x\) approaches 0.
• This is a perfect illustration of how a traditionally problem-causing factor can be bounded through careful analysis.

Oscillating functions are often involved in Squeeze Theorem problems since their unpredictable behavior invites a need for boundary limitations.

Calculus Problem-Solving

Calculus problem-solving is about dissecting complex expressions into simpler, manageable parts. This is exactly what we do by applying the Squeeze Theorem to handle limits involving oscillating functions.

In solving this problem, the steps include:

• Understanding the Objective: Identify that the limit of a function involving oscillations is required as \(x\) approaches a critical point.
• Identifying Boundary Functions: Create functions that bound the oscillating function from above and below, ensuring their limits can be easily computed.
• Applying the Theorem: Verify the conditions of the Squeeze Theorem are met, ensuring that the boundary functions converge to the same limit.
• Visualizing the Problem: Graphing the functions provides a solid visualization, confirming how the central function is squeezed to the desired limit.

This organized approach allows us to tackle complex calculus problems beyond mere algebraic manipulation. It integrates deep understanding of function behavior and limit properties, essential skills for any calculus student.