157 Let N=max{2N1,2N2 +1}. Then |an −ℓ| <ε whenever n≥N. Therefore, limn→∞an =ℓ. This problem is sometimes very helpful to show that a limit exists. For example, consider the sequence defined by x1 =1/2, xn+1 = 1 2+xn for n∈N. We will see later that {x2n+1} and {x2n} both converge to √2−1, so we can conclude that {xn} converges to √2−1. (b) Use a similar method to the solution of part (a). Exercise 2.1.12. Consider the case where ℓ >0. By the definition of limit, we can findn1 ∈Nsuch that |an| > ℓ/2 for all n≥n1. Given any ε >0, we can findn2 ∈Nsuch that |an −ℓ| < ℓε 4 for all n≥n2. Choose n0 =max{n1,n2}. For any n≥n0, one has an+1 an − 1 =| an −an+1| |an| ≤ | an −ℓ| +|an+1 −ℓ| |an| < ℓε 4 + ℓε 4 ℓ 2 =ε. Therefore, limn→∞ an+1 an =1. If ℓ <0, consider the sequence {−an}. The conclusion is no longer true if ℓ =0. A counterexample is an =λn where λ ∈(0,1). SECTION 2.2 Exercise 2.2.3. (a) The limit is calculated as follows: lim n→∞ pn2 +n−n =lim n→∞ √n2 +n−n √n2 +n+n √n2 +n+n =lim n→∞ n √n2 +n+n =lim n→∞ n pn2(1+1/n)+n =lim n→∞ 1 p1+1/n+1 =1/2.
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