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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
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
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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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