Problem 48 - Calculus: Early Transcendentals

Question

Let \( g(x) = \text{sgn}(\sin x) \). (a) Find each of the following limits or explain why it does not exist. (i) \( \displaystyle \lim_{x \to 0^+}g(x) \) (ii) \( \displaystyle \lim_{x \to 0^-}g(x) \) (iii) \( \displaystyle \lim_{x \to 0}g(x) \) (iv) \( \displaystyle \lim_{x \to \pi^+}g(x) \) (v) \( \displaystyle \lim_{x \to \pi^-}g(x) \) (vi) \( \displaystyle \lim_{x \to \pi}g(x) \) (b) For which values of \( a \) does \( \displaystyle \lim_{x \to a}g(x) \) not exist? (c) Sketch a graph of \( g \).

Step-by-Step Solution

Answer

Limits at 0 and \( \pi \) don't exist; fails at \( a = k\pi \).

Step 1: Analyze Function Behavior

The function \( g(x) = \text{sgn}(\sin x) \) represents the sign of the sine function. It takes values \( 1 \), \( -1 \), or \( 0 \) depending on whether \( \sin x \) is positive, negative, or zero, respectively.

Step 2: Evaluate One-Sided Limit at 0 from Right

For \( \lim_{x \to 0^+} g(x) \), as \( x \) approaches 0 from the right, \( \sin x > 0 \), so \( g(x) = 1 \). Therefore, \( \lim_{x \to 0^+} g(x) = 1 \).

Step 3: Evaluate One-Sided Limit at 0 from Left

For \( \lim_{x \to 0^-} g(x) \), as \( x \) approaches 0 from the left, \( \sin x < 0 \), so \( g(x) = -1 \). Thus, \( \lim_{x \to 0^-} g(x) = -1 \).

Step 4: Evaluate Two-Sided Limit at 0

Since the one-sided limits at 0 from the right and left are different, \( \lim_{x \to 0} g(x) \) does not exist because 1 \( eq \) -1.

Step 5: Evaluate One-Sided Limit at \( \pi \) from Right

For \( \lim_{x \to \pi^+} g(x) \), as \( x \) approaches \( \pi \) from the right, \( \sin x < 0 \), so \( g(x) = -1 \). Therefore, \( \lim_{x \to \pi^+} g(x) = -1 \).

Step 6: Evaluate One-Sided Limit at \( \pi \) from Left

For \( \lim_{x \to \pi^-} g(x) \), as \( x \) approaches \( \pi \) from the left, \( \sin x > 0 \), so \( g(x) = 1 \). Thus, \( \lim_{x \to \pi^-} g(x) = 1 \).

Step 7: Evaluate Two-Sided Limit at \( \pi \)

Since the one-sided limits at \( \pi \) from the right and left are different, \( \lim_{x \to \pi} g(x) \) does not exist because -1 \( eq \) 1.

Step 8: Determine Values Where Limit Does Not Exist

The limit \( \lim_{x \to a} g(x) \) does not exist at points where \( \sin(a) = 0 \) since it implies a change in sign. These points are of the form \( a = k\pi \), where \( k \) is an integer.

Step 9: Sketch Graph of g

The graph of \( g(x) \) is composed of horizontal lines at \( y=1 \) when \( \sin x > 0 \), \( y=-1 \) when \( \sin x < 0 \) and points at \( (k\pi, 0) \) as defined by \( \sin(k\pi) = 0 \). This results in a step function appearing infinite but jumping between \( 1 \) and \( -1 \) at \( k\pi \).

Key Concepts

Sign Function

The sign function, often represented as \( \text{sgn}(x) \), is a mathematical operation that reveals the sign of a number or an expression. In simple terms, it tells us whether a value is positive, negative, or zero. For any real number \( x \):

• \( \text{sgn}(x) = 1 \) if \( x > 0 \)
• \( \text{sgn}(x) = -1 \) if \( x < 0 \)
• \( \text{sgn}(x) = 0 \) if \( x = 0 \)

When the sign function is applied to another function, like \( \sin(x) \) in this exercise, it interprets the sign of the sine value at given points. For example, \( \text{sgn}(\sin x) \) will be:

• 1 when \( \sin x > 0 \)
• -1 when \( \sin x < 0 \)
• 0 when \( \sin x = 0 \)

This function forms the basis for understanding the limits in the problem, as it dictates how \( g(x) \) behaves around certain key points such as 0 and \( \pi \). The behavior changes instantly, jumping between 1 and -1, wherever the sine function crosses zero.

Sine Function

The sine function, identified as \( \sin(x) \), is a fundamental trigonometric function representing the y-coordinate of a point on the unit circle as it sweeps around from the positive x-axis. Its range is between -1 and 1, making its application in the sign function quite interesting. As the angle \( x \) varies:

• \( \sin(x) \) is positive in the first and second quadrants.
• It is negative in the third and fourth quadrants.
• It exactly equals zero for angles that are integer multiples of \( \pi \).

With this periodic behavior repeating every \( 2\pi \), it creates intervals where the sign of \( \sin x \) changes. This directly affects the output of \( \text{sgn}(\sin x) \), causing changes at these transition points. These changes significantly involve our understanding of limits and discontinuities in the problem where \( x \) approaches such key angles.

Discontinuity

A discontinuity in a mathematical function refers to points where the function is not continuous, meaning that there is an abrupt change in the function's value. For the function \( g(x) = \text{sgn}(\sin x) \), discontinuities occur at points where the sine function changes sign, which precisely are the multiples of \( \pi \).Discontinuity is observed when:

• The left-hand limit and the right-hand limit around a point differ.
• For \( \lim_{x \to 0}g(x) \), the left-hand limit is -1, and the right-hand limit is 1, hence \( g(x) \) is discontinuous at 0.
• Similarly, at \( \pi \), jumping from 1 to -1 or vice versa also indicates a discontinuity.

Such abrupt changes in values make understanding the limit and continuity principles essential. Wherever \( \sin(x) \) reaches zero, there is a step-like jump in \( g(x) \) from -1 to 1 or vice versa, showing a clear discontinuity.

Piecewise Functions

Piecewise functions are those that have different expressions or rules for different intervals of the domain. The function \( g(x) = \text{sgn}(\sin x) \) is a type of piecewise function. It outputs different constant values based on the sign of \( \sin(x) \). For \( g(x) \), it is defined by:

• \( g(x) = 1 \) when \( \sin x > 0 \)
• \( g(x) = -1 \) when \( \sin x < 0 \)
• \( g(x) = 0 \) at points where \( \sin x = 0 \)

The function effectively resembles a series of steps because it jumps suddenly between 1 and -1 at these zero points. Each piece or segment aligns with intervals formed by \( \sin(x) \) being positive or negative. Such piecewise behavior is crucial in analyzing and sketching the graph of \( g(x) \), which vividly displays its discontinuous nature at integral multiples of \( \pi \). The knowledge of piecewise functions thus aids in comprehending the analysis of limits, especially when they do not exist as in this exercise.