Problem 4
Question
Suppose \(a\) is a constant and \(\delta\) is a positive constant. Give a geometric description of the sets \(\\{x:|x-a|<\delta\\}\) and \(\\{x: 0<|x-a|<\delta\\}\)
Step-by-Step Solution
Verified Answer
Based on the given problem and step-by-step solution, the short answer is:
1. The geometric description of the set \(\\{x: |x-a|<\delta\\}\) on the number line is the open interval (a - 𝛿, a + 𝛿).
2. The geometric description of the set \(\\{x: 0<|x-a|<\delta\\}\) on the number line is the union of two open intervals, (a - 𝛿, a) and (a, a + 𝛿), with the point a removed from both intervals.
1Step 1: Understand the meaning of \(|x-a|\)
The absolute value \(|x-a|\) represents the distance between x and a on the number line. So, in both given sets, we are concerned with distances involving x and a.
2Step 2: Represent the first set on the number line
The first set is \(\\{x: |x-a|<\delta\\}\). This means that all x values have a distance less than 𝛿 from a. Geometrically, this can be represented as an open interval on the number line, going from (a - 𝛿) to (a + 𝛿). It is open because the points (a - 𝛿) and (a + 𝛿) themselves are not included in the set, as x should be strictly closer to a.
In summary, \(\\{x: |x-a|<\delta\\} = (a - \delta, a + \delta)\)
3Step 3: Represent the second set on the number line
The second set is \(\\{x: 0<|x-a|<\delta\\}\). This means that all x values have a distance between 0 and 𝛿 from a. Since the distance cannot be 0, x cannot be equal to a. Geometrically, this can be represented as an open interval on the number line, going from (a - 𝛿) to (a + 𝛿), with the point a removed from the interval, as x should not be equal to a.
In summary, \(\\{x: 0<|x-a|<\delta\\} = (a - \delta, a) \cup (a, a+\delta)\)
These are the geometric descriptions of the given sets, as intervals on the number line.
Key Concepts
Absolute Value InequalitiesOpen IntervalsDistance on Number Line
Absolute Value Inequalities
Absolute value inequalities often describe a range of values rather than a specific one. When we see an expression like \(|x-a| < \delta\), it tells us that the distance between the variable \(x\) and the constant \(a\) is less than \(\delta\).
This is because the absolute value of a number essentially measures how far that number is from zero, which can be thought of as a "distance" measurement. In this context, the inequality units have been shifted to center around \(a\), not zero.
When dealing with inequalities such as \(0 < |x-a| < \delta\), it implies that \(x\) lies within a specified range around \(a\), but without precisely equaling \(a\). The absence of equality tells us that the range is open at both ends.
This is because the absolute value of a number essentially measures how far that number is from zero, which can be thought of as a "distance" measurement. In this context, the inequality units have been shifted to center around \(a\), not zero.
When dealing with inequalities such as \(0 < |x-a| < \delta\), it implies that \(x\) lies within a specified range around \(a\), but without precisely equaling \(a\). The absence of equality tells us that the range is open at both ends.
Open Intervals
An open interval refers to a set of numbers between two endpoints, where the endpoints themselves are not included in the set. This is denoted by parentheses, like \((a, b)\), which represents all numbers \(x\) such that \(a < x < b\).
In the context of our problem, the interval \((a - \delta, a + \delta)\) corresponds to the first inequality \(|x-a|<\delta\). This is a set of all numbers (or distances) less than \(\delta\) from \(a\), illustrating that \(x\) is strictly "inside" the distance boundary around \(a\).
The second set described as \((a - \delta, a) \cup (a, a+\delta)\) includes all points close to \(a\), but specifically excludes \(a\). This highlights how open intervals are perfect for "excluding" specific points while indicating a continuous range.
In the context of our problem, the interval \((a - \delta, a + \delta)\) corresponds to the first inequality \(|x-a|<\delta\). This is a set of all numbers (or distances) less than \(\delta\) from \(a\), illustrating that \(x\) is strictly "inside" the distance boundary around \(a\).
The second set described as \((a - \delta, a) \cup (a, a+\delta)\) includes all points close to \(a\), but specifically excludes \(a\). This highlights how open intervals are perfect for "excluding" specific points while indicating a continuous range.
Distance on Number Line
The number line is a visual representation that we often use to illustrate real numbers and their relationships. Here, distance is a key concept when working with absolute values.
For example, the absolute value \(|x-a|\) can be interpreted as the "distance" between the point \(x\) and the point \(a\) on the number line.
When you want to understand a problem involving expressions like \(|x-a|<\delta\), picture \(a\) as a fixed point, then visualize a "bubble" with radius \(\delta\) around it. The solution set \((a - \delta, a + \delta)\) includes all numbers inside this bubble, but not on its boundary.
In the set \(0<|x-a|<\delta\), this "bubble" still exists, but now the point directly at \(a\) is removed. Grasp the number line as a simple geography of numbers to understand how \(x\) can roam within these bounds, but not to \(a\) itself.
For example, the absolute value \(|x-a|\) can be interpreted as the "distance" between the point \(x\) and the point \(a\) on the number line.
When you want to understand a problem involving expressions like \(|x-a|<\delta\), picture \(a\) as a fixed point, then visualize a "bubble" with radius \(\delta\) around it. The solution set \((a - \delta, a + \delta)\) includes all numbers inside this bubble, but not on its boundary.
In the set \(0<|x-a|<\delta\), this "bubble" still exists, but now the point directly at \(a\) is removed. Grasp the number line as a simple geography of numbers to understand how \(x\) can roam within these bounds, but not to \(a\) itself.
Other exercises in this chapter
Problem 3
The function \(s(t)\) represents the position of an object at time \(t\) moving along a line. Suppose \(s(2)=136\) and \(s(3)=156 .\) Find the average velocity
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Evaluate \(\lim _{x \rightarrow 4}\left(\frac{x^{2}-4 x-1}{3 x-1}\right)\)
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Determine the following limits at infinity. $$\lim _{x \rightarrow-\infty} 3 x^{11}$$
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Consider the function \(F(x)=f(x) / g(x)\) with \(g(a)=0 .\) Does \(F\) necessarily have a vertical asymptote at \(x=a ?\) Explain your reasoning.
View solution