Calculus - limit of the functionsHow do I solve these functions expressed in the image below? Prove the statement using the &, & definition of a limit.x² - 2x - 8x - 4limx →4Given & > 0, we need 8 ---Select----6\x²2x8x - 4= 6< ह(x-4)(x + 2)x - 4x² - 2x - 8 = 6.X-4By the definition of a limit, limX→ 4such that if 0 < x - 4| < |---Select--- ✓-6|--6 < Ex² - 2x - 8x - 4x + 2-6 < ɛ [x #4] ⇒ x - 4| < . Choose & = ---Select--- ✓ |. Then 0 < x - 4| < 8 ⇒ |x- 4| < |---Select--- v⇒Ithen- 6 6 ---Select--- ✓v. We have→x² - 2x - 8x - 4-6---Select--

Answered: Image /qna-images/answer/a8c8a7f4-84e1-48a2-a24d-4fd8d44c8204.jpg

Prove the statement using the , & definition of a limit. x2 - 2x 8 = 6 X-4 lim Given & > 0, we need & ---Select---✓ such that if 0 < x-41 <---Select--- | (x-4) (x + 2) - 6 < x-4 x² - 2x-8 X-4 |x²2x-8 X-4 - 6| < 8 → By the definition of a limit, lim X-4 = 6. , then 1x²2x8. x-4 6 ---Select--- . We have <६ →|x + 2-61 < € [x4] ⇒ |x-41 <&. Choose & = ---Select--- . Then 0 < x-41 → ✓ Then 0 < x 4| < 8 < 8 = |x4| <---Select--- → |x²=2x-8-6 ---Select--- X-4

Prove the statement using the , & definition of a limit. x2 - 2x 8 = 6 X-4 lim Given & > 0, we need & ---Select---✓ such that if 0 < x-41 <---Select--- | (x-4) (x + 2) - 6 < x-4 x² - 2x-8 X-4 |x²2x-8 X-4 - 6| < 8 → By the definition of a limit, lim X-4 = 6. , then 1x²2x8. x-4 6 ---Select--- . We have <६ →|x + 2-61 < € [x4] ⇒ |x-41 <&. Choose & = ---Select--- . Then 0 < x-41 → ✓ Then 0 < x 4| < 8 < 8 = |x4| <---Select--- → |x²=2x-8-6 ---Select--- X-4

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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Question

Calculus - limit of the functions

How do I solve these functions expressed in the image below?

Prove the statement using the &, & definition of a limit.
x² - 2x - 8
x - 4
lim
x →4
Given & > 0, we need 8 ---Select---
-6\
x²2x8
x - 4
= 6
< ह
(x-4)(x + 2)
x - 4
x² - 2x - 8 = 6.
X-4
By the definition of a limit, lim
X→ 4
such that if 0 < x - 4| < |---Select--- ✓
-6|-
-
6 < E
x² - 2x - 8
x - 4
x + 2-6 < ɛ [x #4] ⇒ x - 4| < . Choose & = ---Select--- ✓ |. Then 0 < x - 4| < 8 ⇒ |x- 4| < |---Select--- v
⇒
I
then
- 6 6 ---Select--- ✓
v
. We have
→
x² - 2x - 8
x - 4
-6---Select--

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