Problem 52 - Calculus Early Transcendentals

Question

\(f(x)=\left\\{\begin{array}{ll}{x^{2} \sin \frac{1}{x}} & {\text { if } x \neq 0} \\ {0} & {\text { if } x=0}\end{array}\right.\)

Step-by-Step Solution

Answer

The function is continuous at \( x = 0 \).

Step 1: Identify the problem type

First, recognize that the function described is a piecewise function, which is defined differently at different parts of its domain: \( x^2 \sin\left(\frac{1}{x}\right) \) for \( x eq 0 \) and 0 for \( x = 0 \). The function's behavior at \( x = 0 \) needs special attention.

Step 2: Understand the goal

The task is typically to check the continuity of the function at \( x = 0 \), which often involves finding the limit of the function as \( x \) approaches 0.

Step 3: Calculate the limit as \( x \rightarrow 0 \)

To find \( \lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right) \), note that \( \sin\left(\frac{1}{x}\right) \) is bounded by -1 and 1. Therefore, \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\). Use the Squeeze Theorem to conclude that \( \lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right) = 0 \).

Step 4: Verify continuity at \( x = 0 \)

The function is continuous at \( x = 0 \) if \( \lim_{{x \to 0}} f(x) = f(0) \). Since we found that the limit is 0 and \( f(0) = 0 \), the function is continuous at \( x = 0 \).

Key Concepts

Limit Calculation

Understanding how to calculate limits is vital when dealing with piecewise functions. A limit describes the value that a function approaches as the input comes closer to a particular point. In the given exercise, we are interested in the limit of the function as \( x \) approaches 0 for \( x^2 \sin\left(\frac{1}{x}\right) \).
The key here is recognizing that \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, no matter what \( x \) is. Thus, we frame it as \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\).
This setup allows us to apply the Squeeze Theorem effectively, which ultimately determines that \( \lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right) = 0 \).
Understanding this calculation plays an important part in establishing function behavior and ensures you can accurately determine what the function zeroes in on as \( x \) gets infinitesimally small.

Continuity

Continuity is a property of a function that describes a smooth, unbroken path in its graph. For a function to be continuous at a point, three conditions must be met: the function must be defined at the point, the limit as \( x \) approaches the point must exist, and the value of the limit must equal the function's value at that point.
In our exercise, we confirm the function's continuity at \( x = 0 \) by verifying that:

• The function \( f(x) \) is defined at \( x = 0 \), which gives \( f(0) = 0 \).
• The limit \( \lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right) = 0 \) from our earlier calculation.
• The limit value of 0 equals the function value at \( x = 0 \), \( f(0) = 0 \).

Confirming these points ensures a seamless transition across the point \( x = 0 \), marking the function as continuous at that point. This foundational piece helps to ensure that the function behaves predictably and aligns with real-world applications requiring stability.

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus used to find the limit of a function by comparing it to two other functions whose limits are already known. It is particularly useful when dealing with functions that oscillate or have complex behavior.
The theorem states that if \( g(x) \leq f(x) \leq h(x) \) for all \( x \) in some interval around \( a \), except possibly at \( a \) itself, and \( \lim_{{x \to a}} g(x)=\lim_{{x \to a}} h(x)=L \), then \( \lim_{{x \to a}} f(x)=L \) as well.
In the original function \( x^2 \sin\left(\frac{1}{x}\right) \), the bounds \(-x^2 \) and \( x^2 \) are applied to squeeze the oscillating term \( \sin\left(\frac{1}{x}\right) \) between them.
This effectively 'sandwiches' the behavior of \( x^2 \sin\left(\frac{1}{x}\right) \) and ensures that its limit is zero as \( x \to 0 \), since both boundary functions \( -x^2 \) and \( x^2 \) converge to 0 as \( x \to 0 \).
Utilizing this theorem properly allows students to tackle problems involving complex limits with confidence.