Lesson 7.pdf - 7 Limit Theorems More properties on convergence In order to prove a limit formula it is basic skill to use the definition of limits In

Lesson 7.pdf - 7 Limit Theorems More properties on...

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7. Limit Theorems July 16, 2019 More properties on convergence In order to prove a limit formula, it is basic skill to use the definition of limits. In many cases, we can also use the properties of limits. c Theorem 0.1 Suppose s n s and t n t . Then 1. s n + t n s + t. 2. s n - t n s - t . 3. s n t n st . In particular for any constant k , ks n ks . 4. s n t n s t provided t 6 = 0 . [Example] Before we prove this theorem, we want to illustrate how to use this theorem to prove limits. Let s n = n 2 + n - 1 n 2 - 5 n +4 . Prove lim n →∞ s n = 1. By the theorem above, lim 1 n 2 = lim 1 n · lim 1 n = 0 and lim n →∞ s n = lim n →∞ n 2 + n - 1 n 2 - 5 n + 4 = lim n →∞ 1 + 1 n - 1 n 2 1 - 5 n + 1 n 2 = 1 + lim 1 n - lim 1 n 2 1 - 5 lim 1 n + 4 lim 1 n 2 = 1 + 0 - 0 1 - 0 + 0 = 1 . Proof of Theorem 0.1: 1. For any > 0, we want to find N > 0 such that | ( s n + t n ) - ( s + t ) | < , n > N. 60
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Since s n s , for the , there exists N 1 > 0 such that | s n - s | < 2 , n > N 1 . Since t n t , for the , there exists N 2 > 0 such that | t n - t | < 2 , n > N 2 . Then | ( s n + t n ) - ( s + t ) | = | ( s n - s ) + ( t n - t ) | ≤ | s n - s | + | t n - t | < 2 + 2 = , n > N where N := max { N 1 , N 2 } . Therefore s n + t n s + t . 2. Similar. 3. For any > 0, we want to find N > 0 such that | s n t n - st | < , n > N. Since s n s and t n t , by Theorem 4.1.13, ( s n ) and ( t n ) are bounded, i.e., there exists some M > 0 such that | s n | ≤ M and | t n | ≤ M for all n . This implies | s | ≤ M and | t | ≤ M (see the next theorem). Also, since s n s , for the , there exists N 1 > 0 such that | s n - s | < 2 M , n > N 1 . Since t n t , for the , there exists N 2 > 0 such that | t n - t | < 2 M , n > N 2 . Then for any n > N := max { N 1 , N 2 } , we have | s n t n - st | = | s n ( t n - t ) + ( s n - s ) t | = | s n || t n - t | + | s n - s || t | < M · 2 M + 2 M · M = . Therefore s n t n st . 4. Similar. 61
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Theorem 0.2 Suppose s n s with s n 0 for all n . Then s 0 . Proof: For any > 0, there exists N > 0 such that | s n - s | < , n > N, i.e., - < s n - s < , n > N, which implies s n - < s, n > N.
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  • Fall '08
  • Staff
  • lim, Limit of a function, Limit of a sequence, Suppose sn

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