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# B if the sequence yn 1n for all positive integers n

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Unformatted text preview: satisfy the integer definition. PROVE: If X n = 2n − 1 , then limn → ∞ X n = 2 . n Answer. We begin by computing the "error": € | €n – A| = | 2 – (1/n) – 2| = 1/n. X Then given an ε > 0, the goal is to make this error < ε , so in particular we need (1/n) < ε . Bu t this is true if 1/ε < n. Then for this ε > 0, let N = 1/ε , then for any n > N, 1/n < 1/N, so | Xn – A| = 1/n < 1/N < ε. Thus the definition is satisfied Problem 9 6: Suppose that Zn is a sequence for which this is true: There is a positive integer N such that for every ε > 0, for all n > N, | Zn – A| < ε. What would an example of such a Zn be? What can you prove about Zn that must be true. Answer. This definition says that for all n > N that for that particular n, | Zn – A| < ε for any ε > 0. But this means that it must be true that | Zn – A| ≤ 0, for if the number were positive, there would be a positive ε that would be smaller. But also | Zn – A| ≥ 0 since it is an absolute value,...
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