Example 3 illustrates that direct proofs of even rather simple limits can get

Example 3 illustrates that direct proofs of even

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Example 3 illustrates that direct proofs of even rather simple limits can get complicated. With the limit theorems of §9 we would just write lim 4 n 3 + 3 n n 3 6 lim 4 + 3 n 2 1 6 n 3 lim 4 + 3 · lim( 1 n 2 ) lim 1 6 · lim( 1 n 3 ) 4 . Example 4 Show that the sequence a n ( 1) n does not converge. Discussion . We will assume that lim( 1) n a and obtain a con- tradiction. No matter what a is, either 1 or 1 will have distance at least 1 from a . Thus the inequality | ( 1) n a | < 1 will not hold for all large n . Formal Proof Assume that lim( 1) n a for some a R . Letting 1 in the definition of the limit, we see that there exists N such that n > N implies | ( 1) n a | < 1 . By considering both an even and an odd n > N , we see that | 1 a | < 1 and | − 1 a | < 1 .
2. Sequences 40 Now by the Triangle Inequality 3.7 2 | 1 ( 1) | | 1 a + a ( 1) | ≤ | 1 a |+| a ( 1) | < 1 + 1 2 . This absurdity shows that our assumption that lim( 1) n a must be wrong, so the sequence ( 1) n does not converge. Example 5 Let ( s n ) be a sequence of nonnegative real numbers and suppose that s lim s n . Note that s 0; see Exercise 8.9(a). Prove that lim s n s . Discussion . We must consider > 0 and show that there exists N such that n > N implies | s n s | < . This time we cannot expect to obtain N explicitly in terms of be- cause of the general nature of the problem. But we can hope to show such N exists. The trick here is to violate our training in algebra and “irrationalize the denominator”: s n s ( s n s )( s n + s ) s n + s s n s s n + s . Since s n s we will be able to make the numerator small [for large n ]. Unfortunately, if s 0 the denominator will also be small. So we consider two cases. If s > 0, the denominator is bounded below by s and our trick will work: | s n s | ≤ | s n s | s , so we will select N so that | s n s | < s for n > N . Note that N exists, since we can apply the definition of limit to s just as well as to . For s 0, it can be shown directly that lim s n 0 implies lim s n 0; the trick of “irrationalizing the denominator” is not needed in this case. Formal Proof Case I: s > 0. Let > 0. Since lim s n s , there exists N such that n > N implies | s n s | < s .
§8. A Discussion about Proofs 41 Now n > N implies | s n s | | s n s | s n + s | s n s | s < s s . Case II: s 0. This case is left to Exercise 8.3. Example 6 Let ( s n ) be a convergent sequence of real numbers such that s n 0 for all n N and lim s n s 0. Prove that inf {| s n | : n N } > 0. Discussion . The idea is that “most” of the terms s n are close to s and hence not close to 0. More explicitly, “most” of the terms s n are within 1 2 | s | of s , hence most s n satisfy | s n | ≥ 1 2 | s | . This seems clear from Figure 8.1, but a formal proof will use the triangle inequality.

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• Spring '14
• Limits, Limit of a sequence, Sn, lim sn