Unformatted text preview: to 1 while s n does not converge. 2. Prove that if lim x n = x and if x > 0, then there exists a natural number M such that x n > for all n ≥ M . Set ε = x/ 2. This is positive by assumption. Applying the deﬁnition of convergence, there is M ∈ N so that  x nx  < x/ 2 , ∀ n ≥ M. orx/ 2 < x nx < x/ 2 . The left inequality implies that x n > x/ 2 > 0 for n ≥ M . 1...
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 Spring '10
 MetCalfe
 Calculus, reverse triangle inequality, J. Metcalfe

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