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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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
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
Expert Solution
Step 1: Given

3)lim 3n/(n+1)

4)lim (n-1) /(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.
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.
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