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Problem 71. ldentify all of the theorems implicitly wsed to show that 3n- 100n +1 lim n' (3 - + n' (5+ bn + 4n -7 Notice that this presumes that all of the individual limits ezist. This will become evident as the limit is decomposed.

Problem 71. ldentify all of the theorems implicitly wsed to show that 3n- 100n +1 lim n' (3 - + n' (5+ bn + 4n -7 Notice that this presumes that all of the individual limits ezist. This will become evident as the limit is decomposed.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.1: Infinite Sequences And Summation Notation
Problem 74E
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12:51 1 Safari RealAnalysis-ISBN-fix... CONVERGENCE OF SEQUENCES AND SERIES 82 Just as Theorem 8 says that the limit of a product is the product of the limits, we can prove the analogue for quotients. Theorem 9. Suppose lim a, - a and lim b, - b. Also suppose b 0 and bu 0, Vn. Then lim ) - SCRAPWORK: To prove this, let's look at the special case of trying to prove lim, (+) - t. The general case will follow from this and Theorem Consider t- t - . We are faced with the same dilemma as before; we need to get bounded above. This means we need to get 6 bounded away from zero (at least for large enough n). This can be done as follows. Since b 0, then > 0. Thus, by the definition of lim, b = 6, there exists N1 such that if n> N1, then |6|-||< 6-b< . Thus when n > N1, < b,| and so < t. This says that for n > N1, < 16- bl. We shoukd be able to make this smaller than a given e > 0, provided we make n large enough. END OF SCRAPWORK Problem 70. Prove Theorem 3 These theorems allow us to compute limits of complicated sequences and rigorously verify that these are, in fact, the correct limits without resorting to the definition of a limit. Problem 71. Identify all of the theorems implicitly used to show that 3n - 100n +1 Sn+ 4n? - 7 n (3 - +) 3 - lim (5+-) Notice that this presumes that all of the individual limits ezist. This will become evident as the limit is decomposed. There is one more tool that will prove to be valuable. Theorem 10. (Squeeze Theorem for Sequences) Let (r.). (,), and (t.) be sequences of real numbers with r. S n S ta, V positive integers n. Suppose lim Fn == lim,- tn. Then (s.) must coneerge and lim, -8. A Problem 72. Prove Theorem Id (Hint: This is probably a place where you would want to use s-e<s, <s+e instead of s, - a <e./ The Squeeze Theorem holds even if r. S . St, holds for only sufficiently large n; Le., forn larger than some fixed Ng. This is true because when you find an Ni that works in the original proof, this can be modified by choosing N -max(Ne, N). Also note that this theorem really says two things: (s) converges and it converges to s. This subtle point affects how one should properly use the Squeeze Theorem. CONVERGENCE OF SEQUENCES AND SERIES 83 Example 7. Prove lim -0. pd Proof: Notice that 0< <a = 2. Since lim 0 = 0= lim , then by the Squeeze Theorem, lim -0. Notice that this proof is completely rigorous. Also notice that this is the proper way to use the Squeeze Theorem. Here is an example of an improper use of the Squeeze Theorem. Next 1 Dashboard Calendar To Do Notifications Inbox
Transcribed Image Text:12:51 1 Safari RealAnalysis-ISBN-fix... CONVERGENCE OF SEQUENCES AND SERIES 82 Just as Theorem 8 says that the limit of a product is the product of the limits, we can prove the analogue for quotients. Theorem 9. Suppose lim a, - a and lim b, - b. Also suppose b 0 and bu 0, Vn. Then lim ) - SCRAPWORK: To prove this, let's look at the special case of trying to prove lim, (+) - t. The general case will follow from this and Theorem Consider t- t - . We are faced with the same dilemma as before; we need to get bounded above. This means we need to get 6 bounded away from zero (at least for large enough n). This can be done as follows. Since b 0, then > 0. Thus, by the definition of lim, b = 6, there exists N1 such that if n> N1, then |6|-||< 6-b< . Thus when n > N1, < b,| and so < t. This says that for n > N1, < 16- bl. We shoukd be able to make this smaller than a given e > 0, provided we make n large enough. END OF SCRAPWORK Problem 70. Prove Theorem 3 These theorems allow us to compute limits of complicated sequences and rigorously verify that these are, in fact, the correct limits without resorting to the definition of a limit. Problem 71. Identify all of the theorems implicitly used to show that 3n - 100n +1 Sn+ 4n? - 7 n (3 - +) 3 - lim (5+-) Notice that this presumes that all of the individual limits ezist. This will become evident as the limit is decomposed. There is one more tool that will prove to be valuable. Theorem 10. (Squeeze Theorem for Sequences) Let (r.). (,), and (t.) be sequences of real numbers with r. S n S ta, V positive integers n. Suppose lim Fn == lim,- tn. Then (s.) must coneerge and lim, -8. A Problem 72. Prove Theorem Id (Hint: This is probably a place where you would want to use s-e<s, <s+e instead of s, - a <e./ The Squeeze Theorem holds even if r. S . St, holds for only sufficiently large n; Le., forn larger than some fixed Ng. This is true because when you find an Ni that works in the original proof, this can be modified by choosing N -max(Ne, N). Also note that this theorem really says two things: (s) converges and it converges to s. This subtle point affects how one should properly use the Squeeze Theorem. CONVERGENCE OF SEQUENCES AND SERIES 83 Example 7. Prove lim -0. pd Proof: Notice that 0< <a = 2. Since lim 0 = 0= lim , then by the Squeeze Theorem, lim -0. Notice that this proof is completely rigorous. Also notice that this is the proper way to use the Squeeze Theorem. Here is an example of an improper use of the Squeeze Theorem. Next 1 Dashboard Calendar To Do Notifications Inbox
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