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 defned by x1 = 1/2, 1 xn+1 = for n ∈ N. 2 + xn √ 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 defnition of limit, we can fnd n1 ∈ N such that |an| > ℓ/2 for all n ≥ n1. Given any ε > 0, we can fnd n2 ∈ N such that ℓε |an − ℓ| < for all n ≥ n2. 4 Choose n0 = max{n1,n2}. For any n ≥ n0, one has ℓε ℓε an+1 |an − an+1| |an − ℓ| + |an+1 − ℓ| 4 + 4 − 1 = ≤ < = ε. ℓ an |an| |an| 2 an+1 Therefore, limn→∞ = 1. If ℓ< 0, consider the sequence {−an}. an The conclusion is no longer true if ℓ = 0. A counterexample is an = λ n where λ ∈ (0,1). SECTION 2.2 Exercise 2.2.3. (a) The limit is calculated as follows: √ √ p n2 + n − n n2 + n + n lim n2 + n− n = lim √ n→∞ n→∞ n2 + n + n n = lim √ n→∞ n2 + n + n n = lim p n→∞ n2(1+ 1/n)+ n 1 = lim p = 1/2. n→∞ 1+ 1/n + 1
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