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Prove the statement using the c, & definition of a limit. Given > 0, we need 8 ---Select-- such that if 0 < x-31 < 6, then (4+x)-5--Select--. But (4+x)-5 < x-1| < |||×-31 < x-31 <--Select--. So if we choose 6 ---Select-- then 0 < x-31 <8 (4+x)-5/

Prove the statement using the c, & definition of a limit. Given > 0, we need 8 ---Select-- such that if 0 < x-31 < 6, then (4+x)-5--Select--. But (4+x)-5 < x-1| < |||×-31 < x-31 <--Select--. So if we choose 6 ---Select-- then 0 < x-31 <8 (4+x)-5/

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Prove the statement using the &, & definition of a limit. (4. —×). + lim X→ 3 Given & > 0, we by the definition of a limit. need 8 ---Select---✓ such that if 0 < |x - 3| < 8, then |(4 + 1/3 x) - 5 O Illustrate with a diagram. y 5+ € = 5 5 5-€ y 5+ € 5 5-E 3-6 3-6 3 3 3+6 3+6 X X O y 5+6 5 5-6 y 5+8 5 5-8 ---Select--- 3-E 3-E But 3 3 4 + 3+ € -x) - 5 | < 3+ € X X <E ↔ |x-1|< <E> 1331 |x3| < |x - 3|< ---Select--- V So if we choose 8 = ---Select--- V then 0 < x - 3| < 8 ⇒ |(4 + 3x) - 5| < E. Thus, lim₂ (4 + x) = = 5 3
Transcribed Image Text:Prove the statement using the &, & definition of a limit. (4. —×). + lim X→ 3 Given & > 0, we by the definition of a limit. need 8 ---Select---✓ such that if 0 < |x - 3| < 8, then |(4 + 1/3 x) - 5 O Illustrate with a diagram. y 5+ € = 5 5 5-€ y 5+ € 5 5-E 3-6 3-6 3 3 3+6 3+6 X X O y 5+6 5 5-6 y 5+8 5 5-8 ---Select--- 3-E 3-E But 3 3 4 + 3+ € -x) - 5 | < 3+ € X X <E ↔ |x-1|< <E> 1331 |x3| < |x - 3|< ---Select--- V So if we choose 8 = ---Select--- V then 0 < x - 3| < 8 ⇒ |(4 + 3x) - 5| < E. Thus, lim₂ (4 + x) = = 5 3
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