Chapter 2: Problem 6
Sketch the interval \((a, b)\) on the \(x\) -axis with the point \(c\) inside. Then find a value of \(\delta > 0\) such that \(a < x < b\) whenever \(0<|x-c|<\delta\). $$a=2.7591, \quad b=3.2391, \quad c=3$$
Short Answer
Expert verified
Choose \(\delta = 0.2391\) to satisfy the conditions.
Step by step solution
01
Understand the Interval
We are given the interval \((a, b)\) where \(a = 2.7591\) and \(b = 3.2391\). The point \(c = 3\) lies within this interval because \(a < c < b\).
02
Draw the Interval on the x-axis
Sketch the number line and mark the points \(a = 2.7591\), \(b = 3.2391\), and \(c = 3\) on it. The interval \((a, b)\) is open, which means it includes all points between \(a\) and \(b\), but not \(a\) and \(b\) themselves. The point \(c = 3\) should be clearly marked inside this interval.
03
Understand the Condition |x-c| < δ
We need to find a positive \(\delta\) such that any \(x\) satisfying \(0 < |x-c| < \delta\) also satisfies \(a < x < b\). This means \(x\) must be inside the interval \((a, b)\), close to \(c = 3\).
04
Calculate δ Based on the Interval
Since \(a < x < b\), we have two conditions: \(x < b\) translates to \(c + \delta < b\), and \(x > a\) translates to \(c - \delta > a\). Solving these give us: 1. \(c + \delta < b\) implies \(3 + \delta < 3.2391\) leads to \(\delta < 0.2391\).2. \(c - \delta > a\) implies \(3 - \delta > 2.7591\) leads to \(\delta < 0.2409\).The smaller value \(\delta = 0.2391\) will satisfy both conditions.
05
Conclude the Well-defined δ
Choose \(\delta = 0.2391\). This value ensures that \(x\) within \(\delta\) distance from \(c = 3\) is always inside the interval \((2.7591, 3.2391)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Intervals
In mathematics, an interval is a set of real numbers that includes all numbers between a pair of endpoints and possibly the endpoints themselves. For example, the interval \((a, b)\) discussed here is an *open interval*.
- Open Interval: Only includes values between the endpoints \(a\) and \(b\), not \(a\) or \(b\) themselves.
- Closed Interval: Includes the endpoints, noted as \([a, b]\).
- Half-open Interval: Includes one endpoint, but not the other, like \([a, b)\) or \((a, b]\).
Limits
The concept of limits is foundational in calculus. It describes the value that a function approaches as the input approaches some point. Limits are used to define derivatives, continuity, and integrals.
- The expression \(\lim_{{x \to c}} f(x) = L\) means as \(x\) gets closer and closer to \(c\), \(f(x)\) approaches \(L\).
- Limits allow us to analyze the behaviors of functions at points where they're not necessarily defined.
- Used in defining derivatives, for instance \(f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}.\)
Epsilon-Delta Definition
The epsilon-delta definition is the formal way of describing limits and continuity rigorously.
- Epsilon (ε) represents how close the function value \(f(x)\) needs to be to \(L\), the expected limit value.
- Delta (δ) represents how close \(x\) needs to be to \(c\), except that \(xeq c\).
- Formally, we say \(\lim_{{x \to c}} f(x) = L\) if for every \(\epsilon > 0\) there exists a \(\delta > 0\) so that whenever \(0 < |x - c| < \delta\), it follows that \(|f(x) - L| < \epsilon\).
Real Analysis
Real Analysis is a branch of mathematics dealing with the set of real numbers and the functions defined on them. It provides deep insights into concepts such as convergence, limits, continuity, differentiation, and integration.
- It involves rigorous proofs and definitions, moving beyond just intuitive ideas.
- Studies sequences, series, and the behavior of functions using a more abstract approach.
- Concepts like continuity and differentiability are core topics explored through precise definitions.