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Intro to Analysis in-class_Part_20

# Intro to Analysis in-class_Part_20 - 3.1 Limits and bounds...

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3.1 Limits and bounds 45 Example 3.1.7. Consider the sequence { x n } = { ( - 1) n (1+ 1 n ) } . We’ve seen that sup x n = 3 2 and inf x n = - 2, but these don’t describe the limiting behavior of the sequence. This sequence has limit points 1 and - 1, so limsup x n = 1 and liminf x n = - 1. Theorem 3.1.16. (i) limsup x j is a limit point of { x j } . (ii) limsup x j is the supremum of the set of limits points of { x j } . Proof of (i). Let y = limsup x j , and start with the case y < . Given n N , we can find K such that k K = | y - sup j>k x j | < 1 n . Since y is finite, this shows sup j>k x j is also finite, hence there is some x satisfying ‘ > k, and | x - sup j>k x j | < 1 n . Together, | x - sup j>k x j | < 2 n . By picking larger indices (say k 1 > k ) we can find another point (say x 1 , 1 > ‘ ) satisfying the same criteria. Hence there are infinitely many, and y is a limit point of the sequence. If y = , then { sup j>k x j } is unbounded above. Thus { x j } is unbounded above, and is a limit point. If y = -∞ , then for any - n , n N , there is K such that k K = sup

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