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Prove the statement using the ɛ, 8 definition of a limit. lim (x2 – 2) = 7 x- -3 Given ɛ > 0, we need 8 --Select--- v such that if 0 < |x - (-3)| < 8, then |(x2 - 2) – 7| --Select--- - or upon simplifying we need |x2 – 9| < ɛ whenever 0 < |x + 3| < 8. Notice that if |x + 3| < 1, then -1 < x + 3 < 1 = -7 < x - 3 < -5 = |x – 3| < -Select-- - . So take 8 = --Select--- . Then 0 < |x + 3| < 8 = |x – 3| < 7 and |x + 3| < ɛ/7, so |(x2 - 2) – 7| = |(x + 3)(x - 3)| = |x + 3||x - 3| < (ɛ/7)(7) = --Select--- - Thus, by the definition of a limit, lim (x2 – 2) =|

Prove the statement using the ɛ, 8 definition of a limit. lim (x2 – 2) = 7 x- -3 Given ɛ > 0, we need 8 --Select--- v such that if 0 < |x - (-3)| < 8, then |(x2 - 2) – 7| --Select--- - or upon simplifying we need |x2 – 9| < ɛ whenever 0 < |x + 3| < 8. Notice that if |x + 3| < 1, then -1 < x + 3 < 1 = -7 < x - 3 < -5 = |x – 3| < -Select-- - . So take 8 = --Select--- . Then 0 < |x + 3| < 8 = |x – 3| < 7 and |x + 3| < ɛ/7, so |(x2 - 2) – 7| = |(x + 3)(x - 3)| = |x + 3||x - 3| < (ɛ/7)(7) = --Select--- - Thus, by the definition of a limit, lim (x2 – 2) =|

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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Prac. 12

Prove the statement using the ɛ, & definition of a limit. lim (x2 - 2) = 7 x- -3 Given ɛ > 0, we need & --Select--- such that if 0 < |x - (-3)| < 8, then |(x2 - 2) – 7| Select--- or upon simplifying we need |x2 – 9| < ɛ whenever 0 < |x + 3| < 8. Notice that if |x + 3| < 1, then -1 < x + 3 < 1 = -7 < x - 3 < -5 = |x - 3| <---Select--- . So take 8 = |(x2 – 2) – 7| = |(x + 3)(x – 3)| = |x + 3||x – 3| < (ɛ/7)(7) = | ---Select--- -Select--- . Then 0 < |x + 3| < 8 = ]x – 3| < 7 and |x + 3| < ɛ/7, so lim (x2 - 2) = x - -3 Thus, by the definition of a limit,
Transcribed Image Text:Prove the statement using the ɛ, & definition of a limit. lim (x2 - 2) = 7 x- -3 Given ɛ > 0, we need & --Select--- such that if 0 < |x - (-3)| < 8, then |(x2 - 2) – 7| Select--- or upon simplifying we need |x2 – 9| < ɛ whenever 0 < |x + 3| < 8. Notice that if |x + 3| < 1, then -1 < x + 3 < 1 = -7 < x - 3 < -5 = |x - 3| <---Select--- . So take 8 = |(x2 – 2) – 7| = |(x + 3)(x – 3)| = |x + 3||x – 3| < (ɛ/7)(7) = | ---Select--- -Select--- . Then 0 < |x + 3| < 8 = ]x – 3| < 7 and |x + 3| < ɛ/7, so lim (x2 - 2) = x - -3 Thus, by the definition of a limit,
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