Problem 50 - Calculus - AP Edition

Question

Show that $$f(x)=\left\\{\begin{array}{ll} x^{2} \sin (1 / x), & x \neq 0 \\ 0, & x=0 \end{array}\right.$$ is continuous and differentiable at \(x=0 .\) Sketch the graph of \(f\) near \(x=0\)

Step-by-Step Solution

Answer

The function is continuous and differentiable at \(x=0\), with a graph showing decreasing oscillations as \(x\to0\).

Step 1: Understand the Definition of Continuity

A function is continuous at a point if the limit of the function as it approaches that point from both sides equals the function's value at that point. In mathematical terms, a function \( f(x) \) is continuous at \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \). For this problem, we need to show that \( \lim_{x \to 0} f(x) = f(0) = 0 \).

Step 2: Evaluate the Limit of the Function as x Approaches 0

Consider \( f(x) = x^2 \sin(1/x) \) for \( x eq 0 \). We need to find \( \lim_{x \to 0} x^2 \sin(1/x) \). Notice that \( \sin(1/x) \) is bounded between -1 and 1. Therefore, \(-x^2 \leq x^2 \sin(1/x) \leq x^2\). As \( x \to 0 \), both \( -x^2 \) and \( x^2 \) approach 0. By the squeeze theorem, \( \lim_{x \to 0} x^2 \sin(1/x) = 0 \).

Step 3: Conclude Continuity at x=0

Since \( \lim_{x \to 0} f(x) = 0 \) and \( f(0) = 0 \), the function \( f(x) \) is continuous at \( x = 0 \).

Step 4: Understand the Definition of Differentiability

A function is differentiable at a point if it has a derivative at that point. The derivative \( f'(0) \) is given by \( \lim_{h \to 0} \frac{f(h) - f(0)}{h} \). For differentiability at \( x = 0 \), we need this limit to exist.

Step 5: Evaluate the Derivative at x=0

Calculate \( \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) \). Using the fact that \( \sin(1/h) \) is bounded, we know \(-h \leq h \sin(1/h) \leq h\). As \( h \to 0 \), both \( -h \) and \( h \) approach 0. By the squeeze theorem, \( \lim_{h \to 0} h \sin(1/h) = 0 \). Thus, \( f'(0) = 0 \).

Step 6: Conclude Differentiability at x=0

Since the derivative \( f'(0) \) exists and equals 0, the function \( f(x) \) is differentiable at \( x = 0 \).

Step 7: Sketch the Graph Near x=0

The graph of \( y = f(x) = x^2 \sin(1/x) \) for \( x eq 0 \) oscillates rapidly as \( x \) approaches 0, but the amplitude of the oscillations decreases due to the \( x^2 \) factor. At \( x = 0 \), the graph is flat because \( f(0) = 0 \). The behavior near zero shows increasingly smaller oscillations around the origin.

Key Concepts

Squeeze Theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is a fundamental tool in calculus that helps us determine the limit of a function. It is particularly useful in cases where direct evaluation of the limit is challenging due to the presence of oscillating functions.
To apply the Squeeze Theorem, we need three functions: let's call them \( f(x) \), \( g(x) \), and \( h(x) \). If we know that \( f(x) \leq g(x) \leq h(x) \) for all \( x \) in some interval around \( a \) (excluding possibly at \( a \) itself) and if \( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} g(x) = L \) as well.
In this exercise, to determine the limit of \( f(x) = x^2 \sin(1/x) \) as \( x \to 0 \), we use the fact that \( -1 \leq \sin(1/x) \leq 1 \), giving us the squeeze \(-x^2 \leq x^2 \sin(1/x) \leq x^2\). Both \(-x^2\) and \(x^2\) limit to 0 as we approach 0. Therefore, the middle term, \(x^2 \sin(1/x)\), also limits to 0, demonstrating the power of the Squeeze Theorem.

Definition of Continuity

Continuity at a point is all about the behavior of a function as it approaches that point. A function is continuous at a point \( x = a \) if the following condition is met:

• The function is defined at \( a \), meaning \( f(a) \) is a known value.
• The limit of \( f(x) \) as \( x \) approaches \( a \) exists and equals \( f(a) \), or mathematically \( \lim_{x \to a} f(x) = f(a) \).

To check continuity of the given function \( f(x) \) at \( x = 0 \), we need to show that \( \lim_{x \to 0} f(x) = 0 \) is equal to \( f(0) \). From the original step-by-step solution, since \( \lim_{x \to 0} x^2 \sin(1/x) = 0 \) by using the Squeeze Theorem, and given \( f(0) = 0 \), the function is continuous at \( x = 0 \). This ensures that there are no jumps, breaks, or holes in the graph of the function at that point.

Definition of Differentiability

Differentiability is essentially about a function having a defined tangent slope at a point. A function is differentiable at \( x = a \) if the following is true:

• The derivative \( f'(a) \) exists, which is given by \( \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \).
• If this limit exists, then the function is smooth and has no corners or cusps at \( a \).

For our function \( f(x) \) at \( x = 0 \), we need \( f'(0) \), which is\[\lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h). \]Using the bounded nature of \( \sin(1/h) \) again, with \(-1 \leq \sin(1/h) \leq 1\), we squeeze \(-h \leq h \sin(1/h) \leq h\). As \( h \to 0 \), both bounds approach 0, thus \( \lim_{h \to 0} h \sin(1/h) = 0 \). Therefore, the function is differentiable at \( x = 0 \) with \( f'(0) = 0 \). This confirms not only the smoothness at the origin but also the absence of any sharp turns on the graph for \( f(x) \).