(3) It is a true fact that lim 142 5x – 7 The e-d proof for this limit given below contains a significant flaw in its reasoning. Find and explain the flaw, and then write a correct e-d proof for this limit. Incorrect proof: Let e > 0 be an arbitrary positive number, and let å = min{1, e}. Let r be a real number, and assume that 0 < |r – 2| < 8. Since 8 < 1, our assumption in particular means that 0 < |r – 2| < 1. From the latter inequality, we have: |r – 2| <1 = -1 1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Question
3
(3) It is a true fact that lim
= 1.
1+2 5x – 7
The e-d proof for this limit given below contains a significant flaw in its reasoning. Find and
explain the flaw, and then write a correct e-d proof for this limit.
Incorrect proof: Let e > 0 be an arbitrary positive number, and let å = min{1, ?e}. Let x be
a real number, and assume that 0 < |x – 2| < 8.
Since 8 < 1, our assumption in particular means that 0 < |r – 2| < 1. From the latter
inequality, we have:
|r – 2| <1 = -1< x – 2 < 1
> 1<r< 3
= 5< 5x < 15
> -2 < 5r – 7< 8
2 < [5x – 7| < 8
1
1
1
8
|5x – 72
1
To summarize, our assumption that |æ – 2| < 8 implies that
5x – 72
Since 8 <e, our assumption also implies that |r – 2| < e. Using these two facts together,
we have:
3
|3 — (5х — 7)
- 7
5x – 7
-
10 – 5x
5x – 7
= 5 |x – 2|
< 5 |r – 2|
(by the computation above)
2
1
< 5.
5 2
(since |r – 2| < e)
Transcribed Image Text:3 (3) It is a true fact that lim = 1. 1+2 5x – 7 The e-d proof for this limit given below contains a significant flaw in its reasoning. Find and explain the flaw, and then write a correct e-d proof for this limit. Incorrect proof: Let e > 0 be an arbitrary positive number, and let å = min{1, ?e}. Let x be a real number, and assume that 0 < |x – 2| < 8. Since 8 < 1, our assumption in particular means that 0 < |r – 2| < 1. From the latter inequality, we have: |r – 2| <1 = -1< x – 2 < 1 > 1<r< 3 = 5< 5x < 15 > -2 < 5r – 7< 8 2 < [5x – 7| < 8 1 1 1 8 |5x – 72 1 To summarize, our assumption that |æ – 2| < 8 implies that 5x – 72 Since 8 <e, our assumption also implies that |r – 2| < e. Using these two facts together, we have: 3 |3 — (5х — 7) - 7 5x – 7 - 10 – 5x 5x – 7 = 5 |x – 2| < 5 |r – 2| (by the computation above) 2 1 < 5. 5 2 (since |r – 2| < e)
Expert Solution
Step 1

We have to find the incorrect step in the εδ proof of the given limit which is

limx2 35x-7=1.

First observe the steps of the inequality x-2<1. From this inequality, we 

have to find the condition for 15x-7.

steps

Step by step

Solved in 3 steps

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