2n+1 In Section 8, Example 2, if you follow the same procedure for the result is Зп-5 N = + ??? 9e 3 What integer should replace ????
2n+1 In Section 8, Example 2, if you follow the same procedure for the result is Зп-5 N = + ??? 9e 3 What integer should replace ????
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
What number is ???
![2n+1
In Section 8, Example 2, if you follow the same procedure for
the result is
3n-5'
???
N
9€
What integer should replace ????](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5f558a7-14fc-4024-84d6-4debb1adc6f6%2F3f322828-f684-4667-9a9b-94c2010bf8e9%2F9mtosq8va_processed.png&w=3840&q=75)
Transcribed Image Text:2n+1
In Section 8, Example 2, if you follow the same procedure for
the result is
3n-5'
???
N
9€
What integer should replace ????
![Prove lim n+ = .
Example 2
Зп+1
7n-4
40 2. Sequences
Discussion. For each e > 0, we need to to decide how big n must
be to guarantee n - 2< e. Thus we want
7n-4
21n + 7 – 21n + 12
7(4n – 4)
19
< E.
7(7n – 4)
or
Since 7n – 4 > 0, we can drop the absolute value and manipulate
the inequality further to "solve" for n:
19
< 7n – 4
7€
19
+ 4 < 7n
7€
4
< n.
19
or
or
49€
Our steps are reversible, so we will put N =
could have chosen N to be any number larger than
19
49€
+. Incidentally, we
19
49€
Formal Proof
19
Let e > 0 and let N
+. Then n > N implies n >
49€
19
49€
19
7e
19
hence
7e )
19
hence 7n
+ 4, hence 7n – 4 >
< €, and hence
|
7(7n-4)
Зп+1
7n-4
- < €. This proves lim
Зп+1
7n-4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5f558a7-14fc-4024-84d6-4debb1adc6f6%2F3f322828-f684-4667-9a9b-94c2010bf8e9%2Frd0rq9r_processed.png&w=3840&q=75)
Transcribed Image Text:Prove lim n+ = .
Example 2
Зп+1
7n-4
40 2. Sequences
Discussion. For each e > 0, we need to to decide how big n must
be to guarantee n - 2< e. Thus we want
7n-4
21n + 7 – 21n + 12
7(4n – 4)
19
< E.
7(7n – 4)
or
Since 7n – 4 > 0, we can drop the absolute value and manipulate
the inequality further to "solve" for n:
19
< 7n – 4
7€
19
+ 4 < 7n
7€
4
< n.
19
or
or
49€
Our steps are reversible, so we will put N =
could have chosen N to be any number larger than
19
49€
+. Incidentally, we
19
49€
Formal Proof
19
Let e > 0 and let N
+. Then n > N implies n >
49€
19
49€
19
7e
19
hence
7e )
19
hence 7n
+ 4, hence 7n – 4 >
< €, and hence
|
7(7n-4)
Зп+1
7n-4
- < €. This proves lim
Зп+1
7n-4
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