Chapter 2: Problem 19
Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. \(f(x)=\sqrt{19-x}, \quad L=3, \quad c=10, \quad \epsilon=1\)
Short Answer
Step by step solution
Understand the Problem
Set the Inequality
Solve the First Inequality
Solve the Second Inequality
Combine Results
Find \(\delta\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Real Analysis
- It employs logical reasoning and structures to prepare students for more complex mathematical theories.
- Core elements include the study of limits, continuity, and the real number line.
- It requires careful attention to detail to ensure accuracy in arguments and proofs.
Limits
- A function limit at a point helps us understand the behavior of the function near that point.
- Symbolically, we express the limit of a function as \(x\) approaches \(c\) by \( \lim_{{x \to c}} f(x) = L \).
- The formal epsilon-delta definition provides a precise method for proving limits.
Continuous Functions
- Formally, a function \( f(x) \) is continuous at a point \( c \) if \( \lim_{{x \to c}} f(x) = f(c) \).
- This concept generalizes over intervals; a function is continuous over an interval if it is continuous at every point in that interval.
- Continuous functions have several appealing properties: they are predictable, and small changes in input produce small changes in output.
Mathematical Proofs
- Proofs can be direct, where the conclusion follows straightforwardly from premises, or indirect, where contradiction or contraposition are used.
- Understanding mathematical proofs requires a grasp of formal logic, mathematical language, and symbolic representation.
- Epsilon-delta proofs in limits and continuity use this precise language to prove statements about function behavior.