Problem 73 - Calculus Early Transcendentals

Question

Let $$ f(x)=\left\\{\begin{array}{ll}{x^{2} \sin \frac{1}{x},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right. $$ (a) Show that \(f\) is continuous at \(x=0\). (b) Use Definition 2.2 .1 to show that \(f^{\prime}(0)\) does not exist. (c) Find \(f^{\prime}(x)\) for \(x \neq 0\). (d) Determine whether \(\lim _{x \rightarrow 0} f^{\prime}(x)\) exists.

Step-by-Step Solution

Answer

(a) Continuous at 0. (b) \(f'(0)\) does not exist. (c) \(f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}\). (d) Limit does not exist.

Step 1: Show Continuity at x=0

To show that the function is continuous at \( x=0 \), we need to verify three conditions: \( f(0) \) is defined, \( f(x) \) has a limit as \( x \to 0 \), and \( \lim_{x \to 0} f(x) = f(0) \).1. \( f(0) = 0 \)2. As \( x \to 0 \), the expression becomes \( x^2 \sin \frac{1}{x} \). Since \( -1 \leq \sin \frac{1}{x} \leq 1 \), we have \( -x^2 \leq x^2 \sin \frac{1}{x} \leq x^2 \), so by the Squeeze Theorem, \( \lim_{x \to 0} x^2 \sin \frac{1}{x} = 0 \).3. Thus, \( \lim_{x \to 0} f(x) = f(0) \).Therefore, \( f(x) \) is continuous at \( x=0 \).

Step 2: Show f'(0) Does Not Exist

To show \( f'(0) \) does not exist, use the definition of the derivative: \( f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x} = \lim_{x \to 0} x \sin \frac{1}{x} \).The limit \( \lim_{x \to 0} x \sin \frac{1}{x} \) does not exist, as the term \( \sin \frac{1}{x} \) oscillates between \(-1\) and \(1\). Therefore, the limit is undefined and \( f'(0) \) does not exist.

Step 3: Find f'(x) for x ≠ 0

For \( x eq 0 \), apply the product and chain rules:\( f(x) = x^2 \sin \frac{1}{x} \).The derivative is:\[ f'(x) = \frac{d}{dx}(x^2) \cdot \sin \frac{1}{x} + x^2 \cdot \frac{d}{dx}(\sin \frac{1}{x}) \]\[ = 2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} \cdot \left(-\frac{1}{x^2}\right) \]\[ = 2x \sin \frac{1}{x} - \cos \frac{1}{x} \]Thus, \( f'(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x} \) for \( x eq 0 \).

Step 4: Determine Limit of f'(x) as x → 0

Consider \( \lim_{x \to 0} f'(x) = \lim_{x \to 0} (2x \sin \frac{1}{x} - \cos \frac{1}{x}) \).The term \( 2x \sin \frac{1}{x} \to 0 \) as \( x \to 0 \). However, \( \cos \frac{1}{x} \) oscillates between \(-1\) and \(1\), so the limit of \( f'(x) \) does not exist as \( x \to 0 \).

Key Concepts

Limit of a Function

The concept of a limit is crucial in calculus and helps us understand the behavior of functions as they approach specific values. In simple terms, the limit of a function is what the function values get closer to as the input (or 'x' value) approaches a certain number. For example, if we consider the function given in the exercise, we want to find what value the function approaches as x gets closer to zero.
To determine if the function is continuous at a point, like x=0 in this case, we look at three things: whether the function is defined at that point, whether a limit exists as x approaches the point, and whether the limit is equal to the function's value at that point.
In our example, the limit of the function as x approaches 0 is zero, thanks to the Squeeze Theorem. This allows us to conclude that the function is indeed continuous at x=0.

Squeeze Theorem

The Squeeze Theorem is a powerful tool when it comes to dealing with difficult limits. It is particularly useful when the function you're working with has a known bounded component, like sine or cosine, between two other functions whose limits are easy to calculate.
In this scenario, the function is multiplied by another function (like x²). Because \(-1 \leq \sin(1/x) \leq 1\) for all x, we can "squeeze" our function between -x² and x².
Consequently, as x approaches zero, both bounds (-x² and x²) approach 0. Thus, according to the Squeeze Theorem, the limit of x²\(\sin(1/x)\) as x approaches zero is also zero. This application of the Squeeze Theorem helped in proving the continuity of the function at x=0 in the step-by-step solution above.

Definition of Derivative

The derivative of a function gives us the rate at which the function's value changes as its input changes. In basic terms, it tells us how steep a function's graph is at a given point.
The fundamental definition of a derivative at a point, say x=0, involves evaluating the limit: \( f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} \). This expression is simply the slope of the tangent line to the graph of the function at x=0.
In our problem, as x approaches zero, the primary function \( x\sin(1/x) \) oscillates wildly between two extreme values because of the sine component. This oscillation makes it impossible to pin down a single slope at x=0, which is why \( f'(0) \), the derivative at x=0, does not exist. The concept of an oscillating function plays a big role here.

Oscillating Functions

Oscillating functions are functions that swing back and forth between certain values. A typical example is the sine function, which oscillates between -1 and 1. Oscillating behavior can complicate the process of finding limits or derivatives because it can prevent the function from settling towards a single value.
In the exercise mentioned, the component \(\sin(1/x)\) is oscillating infinitely as x approaches zero. This infinite oscillation is impossible to capture with a single value, causing challenges in determining certain limits or derivatives.
Particularly, when searching for the derivative at x=0, the oscillation of \(\sin(1/x)\) makes the expression \(x\sin(1/x)\) unpredictable as x approaches zero. This unpredictability leads to the conclusion that the derivative doesn't exist at that particular point. Oscillating functions show how complex and intriguing calculus can be.