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Prove the statement using the , & definition of a limit. lim 9 + 3x = 3 Given e > 0, we need 6 ---Select-- such that if 0 < |x-11 < 6, then 9 + 3x - 3|--Select--. But | 9 + 3x - 3| < | ³x - ³| < | ||× - 11 < £ |× - 11 < --Select--. So if we choose &=-- Select, then 0 < x-11 < 8 (9 + 3x) - 3|< e. Thus, lim (9 + 3x) = 3 by the definition of a limit.

Prove the statement using the , & definition of a limit. lim 9 + 3x = 3 Given e > 0, we need 6 ---Select-- such that if 0 < |x-11 < 6, then 9 + 3x - 3|--Select--. But | 9 + 3x - 3| < | ³x - ³| < | ||× - 11 < £ |× - 11 < --Select--. So if we choose &=-- Select, then 0 < x-11 < 8 (9 + 3x) - 3|< e. Thus, lim (9 + 3x) = 3 by the definition of a limit.

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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Q12. Please answer all the parts to this question 

Prove the statement using the &, & definition of a limit. 9 + 3x 4 lim x → 1 Given & > 0, we need definition of a limit. = 3 --Select--- such that if 0 < x - 1| < 6, then 9 + 3x 4 -Select--- V . But 9 + 3x 4 - 3| 3 < E 3x - 3 | - < E 12/11 ||x − 1| < ɛ ⇒ |x-1| <---Select--- V So if we choose > = 9 3x -Select--- V then 0 |x-1| < 8⇒ |( ⁹ + ³x) - 3| 4 - 3 < E. Thus, lim x → 1 9 + 3x 3 by the
Transcribed Image Text:Prove the statement using the &, & definition of a limit. 9 + 3x 4 lim x → 1 Given & > 0, we need definition of a limit. = 3 --Select--- such that if 0 < x - 1| < 6, then 9 + 3x 4 -Select--- V . But 9 + 3x 4 - 3| 3 < E 3x - 3 | - < E 12/11 ||x − 1| < ɛ ⇒ |x-1| <---Select--- V So if we choose > = 9 3x -Select--- V then 0 |x-1| < 8⇒ |( ⁹ + ³x) - 3| 4 - 3 < E. Thus, lim x → 1 9 + 3x 3 by the
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