Problem 60 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative \( f'(0) \) exists and is equal to 0.

Step 1: Define the Derivative at Zero

To find if the derivative \( f'(0) \) exists, we need to evaluate the limit: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Since \( f(0) = 0 \) as given, the expression simplifies to: \[ f'(0) = \lim_{h \to 0} \frac{h^2 \sin \frac{1}{h}}{h} \] This reduces to: \[ \lim_{h \to 0} h \sin \frac{1}{h} \]

Step 2: Evaluate the Simplified Limit

To determine this limit, note that \( \sin \frac{1}{h} \) oscillates between -1 and 1 for all \( h eq 0 \). Therefore, the whole expression is bounded by: \[ -|h| \leq h \sin \frac{1}{h} \leq |h| \] As \( h \to 0 \), both \( -|h| \) and \( |h| \) approach 0.

Step 3: Apply the Squeeze Theorem

Since \( h \sin \frac{1}{h} \) is squeezed between \( -|h| \) and \( |h| \), and both these bounds approach 0 as \( h \to 0 \), we can apply the Squeeze Theorem. Thus, \[ \lim_{h \to 0} h \sin \frac{1}{h} = 0 \]

Step 4: Conclusion for the Derivative

Since the limit evaluated in the previous step results in 0, we find that: \[ f'(0) = 0 \] Therefore, the derivative \( f'(0) \) exists and is equal to 0.

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a vital concept in calculus used to evaluate limits of functions that are difficult to analyze directly. It states that if a function is 'squeezed' between two other functions that have the same limit at a certain point, then the function in question will also have that same limit at that point. In more formal terms, if \( g(x) \leq f(x) \leq h(x) \) near \( a \), and \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} f(x) = L \).
This theorem is particularly useful when dealing with functions involving oscillations, such as trigonometric functions, where pinpointing the behavior of the function directly can be challenging. By using bounding functions, we can "squeeze" the troublesome function and determine its limit precisely, as seen in evaluating \( \lim_{h \to 0} h \sin \frac{1}{h} \) in this exercise.

trigonometric functions

Trigonometric functions like sine and cosine often appear in calculus, particularly in limit evaluations. They are periodic and oscillate within specific ranges. For example, \( \sin(x) \) oscillates between -1 and 1 for all real numbers.
If a trigonometric function appears in a limit problem, it often requires careful handling because of its oscillatory nature. This is exactly what happens with \( \sin \frac{1}{x} \); it oscillates infinitely as \( x \to 0 \), making it hard to evaluate directly.
In this exercise, despite the oscillation, we take advantage of the fact that its amplitude is bounded, which allows the Squeeze Theorem to come into play. This method efficiently handles the challenge posed by trigonometric oscillations.

limit evaluation

Limit evaluation is a fundamental skill in calculus, and it involves finding the value that a function approaches as the input approaches a certain point. This requires understanding the behavior of functions near particular points, including the effects of continuity or discontinuity.
In this exercise, to find whether the derivative \( f'(0) \) exists, we evaluate the limit \( \lim_{h \to 0} h \sin \frac{1}{h} \). A direct approach to this might be difficult due to the oscillating \( \sin \frac{1}{h} \), but by revisiting the properties of limits and bounding the function, the Squeeze Theorem can be applied to find that the limit evaluates to 0.
Limit evaluation thus combines logical reasoning with theorem applications to effectively solve complex problems.

piecewise function

A piecewise function is defined by different expressions based on different intervals of the input variable. This type of function can exhibit different behaviors depending on the value of the input variable, making them quite interesting but also challenging at times when evaluating limits or derivatives.
In this problem, \( f(x) \) is defined as \( x^2 \sin \frac{1}{x} \) when \( x eq 0 \) and as 0 when \( x = 0 \). The challenge was to evaluate the derivative at \( x = 0 \), which often requires careful analysis of both sections of the piecewise function, especially when they involve potentially undefined points like the sine of an inverse.
Piecewise functions are useful in engineering, physics, and various other practical fields, where different rules might apply in different scenarios, but they require a comprehensive understanding and careful handling, much like in the exercise.