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Dashboard Prove the following limit. SOLUTION lim (2x5)= 7 x 6 But |(2x - 5) 71 = |2x - 12| = 2 1. Preliminary analysis of the problem (guessing a value for 8). Let & be a given positive number. We want to find a number 8 such that if 0 < x 61 < & then |(2x - 5) - 71 < E. that is, if 0 < x-61 < 8 if 0 < x 61 < 8 |(2x - 5) 71 = E This suggests that we ould choose 6 = 2 then = 2 then 2. Proof (showing that & works). Given > 0, choose 6 = =E 2 < 28 Comma Practice - Google Docs Therefore, we want & such that If 0 <

Dashboard Prove the following limit. SOLUTION lim (2x5)= 7 x 6 But |(2x - 5) 71 = |2x - 12| = 2 1. Preliminary analysis of the problem (guessing a value for 8). Let & be a given positive number. We want to find a number 8 such that if 0 < x 61 < & then |(2x - 5) - 71 < E. that is, if 0 < x-61 < 8 if 0 < x 61 < 8 |(2x - 5) 71 = E This suggests that we ould choose 6 = 2 then = 2 then 2. Proof (showing that & works). Given > 0, choose 6 = =E 2 < 28 Comma Practice - Google Docs Therefore, we want & such that If 0 <

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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Dashboard Prove the following limit. SOLUTION lim (2x - 5) = 7 X-6 But |(2x5)-7| = |2x - 12| = 2 1. Preliminary analysis of the problem (guessing a value for 8). Let & be a given positive number. We want to find a number & such that if 0 < x-61 < 8 then |(2x - 5) - 71 < E. that is, if 0 < x-61 < 8 if 0 < x 61 < 8 This suggests that we should choose 6 = E 2 |(2x - 5) 71 = then 2. Proof (showing that & works). Given & > 0, choose 6 = = 2 then = E Comma Practice - Google Docs < 28 Therefore, we want & such that If 0 < < E GP G Grammarly < 6, then we get the following.
Transcribed Image Text:Dashboard Prove the following limit. SOLUTION lim (2x - 5) = 7 X-6 But |(2x5)-7| = |2x - 12| = 2 1. Preliminary analysis of the problem (guessing a value for 8). Let & be a given positive number. We want to find a number & such that if 0 < x-61 < 8 then |(2x - 5) - 71 < E. that is, if 0 < x-61 < 8 if 0 < x 61 < 8 This suggests that we should choose 6 = E 2 |(2x - 5) 71 = then 2. Proof (showing that & works). Given & > 0, choose 6 = = 2 then = E Comma Practice - Google Docs < 28 Therefore, we want & such that If 0 < < E GP G Grammarly < 6, then we get the following.
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