Use Definition 4.2.1 (and not the Algebraic Limit Theorem for Functional Limits, or Theorem 4.2.3) to prove the following statements about limits: (d) lim= . x→3
Use Definition 4.2.1 (and not the Algebraic Limit Theorem for Functional Limits, or Theorem 4.2.3) to prove the following statements about limits: (d) lim= . x→3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Definition 4.2.1 (Functional Limit):** Let \( f : A \to \mathbb{R} \), and let \( c \) be a limit point of the domain \( A \). We say that \(\lim_{x \to c} f(x) = L\) provided that, for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x-c| < \delta\) (and \(x \in A\)) it follows that \(|f(x) - L| < \epsilon\).
This is often referred to as the “\(\epsilon\)-\(\delta\) version” of the definition for functional limits. Recall that the statement
\[|f(x) - L| < \epsilon\]
is equivalent to \(f(x) \in V_\epsilon(L)\).
Likewise, the statement
\[|x - c| < \delta\]
is satisfied if and only if \(x \in V_\delta(c)\).
---
**Definition 4.2.1B (Functional Limit: Topological Version):** Let \( c \) be a limit point of the domain of \( f : A \to \mathbb{R} \). We say \(\lim_{x \to c} f(x) = L\) provided that, for every \(\epsilon\)-neighborhood \(V_\epsilon(L)\) of \(L\), there exists a \(\delta\)-neighborhood \(V_\delta(c)\) around \(c\) with the property that for all \(x \in V_\delta(c)\) different from \(c\) (with \(x \in A\)), it follows that \(f(x) \in V_\epsilon(L)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa923b6f-81dd-482c-8885-6de6bc295751%2Fdd6260a9-0be9-4d8c-83df-89e69229043d%2Fktdcb8v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Definition 4.2.1 (Functional Limit):** Let \( f : A \to \mathbb{R} \), and let \( c \) be a limit point of the domain \( A \). We say that \(\lim_{x \to c} f(x) = L\) provided that, for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x-c| < \delta\) (and \(x \in A\)) it follows that \(|f(x) - L| < \epsilon\).
This is often referred to as the “\(\epsilon\)-\(\delta\) version” of the definition for functional limits. Recall that the statement
\[|f(x) - L| < \epsilon\]
is equivalent to \(f(x) \in V_\epsilon(L)\).
Likewise, the statement
\[|x - c| < \delta\]
is satisfied if and only if \(x \in V_\delta(c)\).
---
**Definition 4.2.1B (Functional Limit: Topological Version):** Let \( c \) be a limit point of the domain of \( f : A \to \mathbb{R} \). We say \(\lim_{x \to c} f(x) = L\) provided that, for every \(\epsilon\)-neighborhood \(V_\epsilon(L)\) of \(L\), there exists a \(\delta\)-neighborhood \(V_\delta(c)\) around \(c\) with the property that for all \(x \in V_\delta(c)\) different from \(c\) (with \(x \in A\)), it follows that \(f(x) \in V_\epsilon(L)\).
Transcribed Image Text:Use Definition 4.2.1 (and not the Algebraic Limit Theorem for Functional Limits, or Theorem 4.2.3) to prove the following statements about limits:
(d) \(\lim_{{x \to 3}} \frac{1}{x} = \frac{1}{3}\).
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