Prove the statement using the , & definition of a limit. 6 + 4x lim x 1 = 2 Given & > 0, we need ---Select--- choose 8 = ---Select--- · then 0 6 + 4x ---Select--- ✓ But 6 + 4x 5 5 <&. Thus, lim (6+4x) = 2 by the definition of a limit. such that if 0 < |x-1| < 6, then > | ( 6 + 4x) - 2 | < |x-1| < 8⇒ |54|

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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.
6 + 4x
lim
x → 1 5
= 2
Given & > 0, we need & ---Select---✓ such that if 0 < |x − 1| < 6, then
|(6 + 4x) - 2|
5
choose 6 = ---Select--- V then 0 < |x-1| < 6⇒
6 + 4x
5
2
<&. Thus, lim
X→ 1
6 + 4x
5
= 2 by the definition of a limit.
---Select--- V
6 + 4x
5
■
But
-
2 < E
4x - 4
5
< ε #
4
|x − 1| < ɛ ⇒ |x − 1| < |---Select--- --
■
So if we
Transcribed Image Text:Prove the statement using the &, & definition of a limit. 6 + 4x lim x → 1 5 = 2 Given & > 0, we need & ---Select---✓ such that if 0 < |x − 1| < 6, then |(6 + 4x) - 2| 5 choose 6 = ---Select--- V then 0 < |x-1| < 6⇒ 6 + 4x 5 2 <&. Thus, lim X→ 1 6 + 4x 5 = 2 by the definition of a limit. ---Select--- V 6 + 4x 5 ■ But - 2 < E 4x - 4 5 < ε # 4 |x − 1| < ɛ ⇒ |x − 1| < |---Select--- -- ■ So if we
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