3. Use the definition of convergence to prove lim Ross 9 is allowed. 4. Use the definition of convergence to prove lim Ross 9 is allowed. 3n n+1 n² +1 = 3; no theorem in = = 0; no theorem in
3. Use the definition of convergence to prove lim Ross 9 is allowed. 4. Use the definition of convergence to prove lim Ross 9 is allowed. 3n n+1 n² +1 = 3; no theorem in = = 0; no theorem in
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
![3. Use the definition of convergence to prove lim 3n
Ross 9 is allowed.
n+1
=
3; no theorem in
n-1
4. Use the definition of convergence to prove lim 21 = 0; no theorem in
Ross 9 is allowed.
n²+1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d5e88e6-b1af-4aea-9b08-2dadd85f5e2c%2F50c88832-b16c-4c3c-b7b3-4a3987916ff9%2Fjinw8ju_processed.png&w=3840&q=75)
Transcribed Image Text:3. Use the definition of convergence to prove lim 3n
Ross 9 is allowed.
n+1
=
3; no theorem in
n-1
4. Use the definition of convergence to prove lim 21 = 0; no theorem in
Ross 9 is allowed.
n²+1
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Follow-up Question
can you use the form in picture to proof them?
![Formal Proof
Let e > 0 and let N =
54
max{2, ¹}. Then n > N implies n > € 9
54
27n
hence 2 < €, hence < €. Since n > 2, we have 2³ ≤ n³ – 6 and
n³/2
n²
also 27n3n+24.
Thus n > N implies
and hence
as desired.
lim
3n+24
n³-6
4n³ + 3n"
3
n³-6
<
=
= lim
27n
1/1/n³
3
4n³ + 3n
n³-6
Example 3 illustrates direct proofs of even rather simple limits
can get complicated. With the limit theorems of §9 we would just
write
4+
1
=
n3
54
n²
< €₂
4 < €,
=
lim 4+ 3. lim(2)
lim 1-6. lim(3)
= 4.](https://content.bartleby.com/qna-images/question/1d5e88e6-b1af-4aea-9b08-2dadd85f5e2c/f5830525-fb51-4bf7-9a16-db85d0e5399a/zpoeibn_thumbnail.png)
Transcribed Image Text:Formal Proof
Let e > 0 and let N =
54
max{2, ¹}. Then n > N implies n > € 9
54
27n
hence 2 < €, hence < €. Since n > 2, we have 2³ ≤ n³ – 6 and
n³/2
n²
also 27n3n+24.
Thus n > N implies
and hence
as desired.
lim
3n+24
n³-6
4n³ + 3n"
3
n³-6
<
=
= lim
27n
1/1/n³
3
4n³ + 3n
n³-6
Example 3 illustrates direct proofs of even rather simple limits
can get complicated. With the limit theorems of §9 we would just
write
4+
1
=
n3
54
n²
< €₂
4 < €,
=
lim 4+ 3. lim(2)
lim 1-6. lim(3)
= 4.
Solution
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