Use algebra to find the largest possible value of 8 or small- est possible value of N that makes each implication in Exer- cises 23-28 true. Then verify and support your answers with labeled graphs. 23. If 0 < |x – 2| < 8, then |(3x – 1) – 5| < 0.25. 24. If 0 < |x – 3| < 8, then | 1 1 < 0.2. 25. If x e (1,1+ 8), then |Vx–1– 0| < 0.5. 1 26. If x e (3 – 8,3), then 3 > 1000. - X 27. Ifx > N, then 1 -0 < 0.001. 28. If x > N, then 1– 2x < -500. For each limit statement lim f(x) = L in Exercises 29–40, use algebra to find 8 > 0 in terms of e > 0 so that if 0 < |x – c| < 8, then |f(x) – L| < e. 29. lim(x + 5) = 8 30. lim (4 – 2x) = 8 %3D X-2 31. lim (3 – 4x) = 3 32. lim (3x + 8) = 11 X+1 33. lim (5x' – 1) = -1 34. lim(x' - 6х + 5) %3D — 4 X3
Use algebra to find the largest possible value of 8 or small- est possible value of N that makes each implication in Exer- cises 23-28 true. Then verify and support your answers with labeled graphs. 23. If 0 < |x – 2| < 8, then |(3x – 1) – 5| < 0.25. 24. If 0 < |x – 3| < 8, then | 1 1 < 0.2. 25. If x e (1,1+ 8), then |Vx–1– 0| < 0.5. 1 26. If x e (3 – 8,3), then 3 > 1000. - X 27. Ifx > N, then 1 -0 < 0.001. 28. If x > N, then 1– 2x < -500. For each limit statement lim f(x) = L in Exercises 29–40, use algebra to find 8 > 0 in terms of e > 0 so that if 0 < |x – c| < 8, then |f(x) – L| < e. 29. lim(x + 5) = 8 30. lim (4 – 2x) = 8 %3D X-2 31. lim (3 – 4x) = 3 32. lim (3x + 8) = 11 X+1 33. lim (5x' – 1) = -1 34. lim(x' - 6х + 5) %3D — 4 X3
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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Transcribed Image Text:Use algebra to find the largest possible value of 8 or small-
est possible value of N that makes each implication in Exer-
cises 23-28 true. Then verify and support your answers with
labeled graphs.
23. If 0 < |x – 2| < 8, then |(3x – 1) – 5| < 0.25.
24. If 0 < |x – 3| < 8, then
| 1
1
< 0.2.
25. If x e (1,1+ 8), then |Vx–1– 0| < 0.5.
1
26. If x e (3 – 8,3), then
3
> 1000.
- X
27. Ifx > N, then
1
-0 < 0.001.
28. If x > N, then 1– 2x < -500.
For each limit statement lim f(x) = L in Exercises 29–40, use
algebra to find 8 > 0 in terms of e > 0 so that if 0 <
|x – c| < 8, then |f(x) – L| < e.
29. lim(x + 5) = 8
30. lim (4 – 2x) = 8
%3D
X-2
31. lim (3 – 4x) = 3
32. lim (3x + 8) = 11
X+1
33. lim (5x' – 1) = -1
34. lim(x' - 6х + 5) %3D — 4
X3
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