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 n-x | < x/ 2 , ∀ n ≥ M. or-x/ 2 < x n-x < x/ 2 . The left inequality implies that x n > x/ 2 > 0 for n ≥ M . 1...
View Full Document
- Spring '10
- Calculus, reverse triangle inequality, J. Metcalfe