Problem 16 - Contemporary Precalculus

Question

Explain why the function is not continuous at the given number. $$g(x)=\left\\{\begin{array}{ll} \sin (\pi / x) & \text { if } x \neq 0 \\ 1 & \text { if } x=0 \end{array} \text { at } x=0\right.$$

Step-by-Step Solution

Answer

Answer: The function g(x) is not continuous at x=0 because the limit does not exist as x approaches 0, even though the function is defined at x=0, which means it does not satisfy the condition for continuity at that point.

Step 1: Check if the function is defined at x=0

According to the definition of the function g(x), when x=0, g(x)=1. Therefore g(0) is clearly defined.

Step 2: Find the limit of g(x) as x approaches 0

Now we need to determine if the limit of g(x) exists when x approaches 0. Recall that the limit exists if the limits from both left and right side of the point are equal. We need to calculate the following limits: $$\lim_{x\to 0^-}g(x) \text{ and } \lim_{x\to 0^+}g(x)$$ The function behaves differently for negative and positive x. For x approaching 0 from the left $$\lim_{x\to 0^-}g(x)=\lim_{x\to 0^-}\sin(\pi/x).$$ For x approaching 0 from the right $$\lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}\sin(\pi/x).$$ Since the sine function oscillates between -1 and 1, as x tends to 0, the values inside the sine function will oscillate constantly, making the function oscillate between -1 and 1 as well. Therefore, the limits do not exist, and the function is not continuous at x=0 because it does not satisfy condition 2 of continuity.

Step 3: Conclusion

The function g(x) is not continuous at x=0 because the limit does not exist as x approaches 0, even though the function is defined at x=0. The continuous condition is not satisfied, and we can conclude that g(x) is not continuous at x=0.

Key Concepts

Limits

The concept of limits is fundamental to understanding continuity in functions. A limit helps us define the behavior of a function as it approaches a specific point. To determine if a function is continuous at a point, such as at \( x = 0 \) in our exercise, we need to verify if the limit of the function matches the function's value at that point. For a function to be continuous at a given point:

• The function must be defined at that point.
• The limit of the function as it approaches the point from the left side must equal the limit as it approaches from the right side.
• The limit at that point must equal the function's value at that point.

In our given function \( g(x) \), even though the value of \( g(0) \) is defined as 1, the limits from both sides of 0 are not consistent due to the oscillating nature of \( \sin(\pi/x) \). Thus, the limit does not exist, leading to discontinuity at \( x=0 \).

Piecewise Functions

Piecewise functions are functions defined by different expressions based on the input value. They can look intimidating, but they simply mean that the function behaves differently in different parts of its domain.
In piecewise functions, there are often different rules or formulas for different sections of the x-axis. It's crucial to verify continuity at points where the function definition changes.
For example, in our function \( g(x) \), we have:

• \( g(x) = \sin(\pi/x) \) when \( x eq 0 \)
• \( g(x) = 1 \) when \( x = 0 \)

The challenge arises at the transition point \( x = 0 \), where the behavior of \( \sin(\pi/x) \) does not align smoothly with the value \( g(0) = 1 \). This is where issues of discontinuity typically arise in piecewise functions. The key is whether the transition from one piece to another is seamless in terms of limits.

Trigonometric Functions

Trigonometric functions like sine and cosine have properties that make them oscillate between certain values. Understanding these properties is essential, especially when they appear in piecewise functions.
In our case, the function \( \sin(\pi/x) \) creates intriguing behavior as \( x \) approaches 0. It's crucial to realize:

• The sine and cosine functions oscillate between -1 and 1, regardless of their inputs.
• As \( x \) gets closer to 0, the frequency of the oscillations in \( \sin(\pi/x) \) increases dramatically, getting more and more rapid.
• Due to these rapid oscillations, the function doesn't settle at a particular value, making the limit indefinite as \( x \) approaches 0.

This inherent oscillating property of \( \sin(\pi/x) \) causes the described discontinuity in the piecewise function. Recognizing how trigonometric functions can influence limits and continuity is essential for understanding these unique behaviors.