35 ■ Example 2.1.3 Consider the sequence {an} where 3n2 + 4 an = . 2n2 + n+ 5 We will prove directly from the defnition that this sequence converges to a = 3/2. Let ε > 0. We frst search for a suitable positive integer N. To that end, we simplify and estimate the expression |an − a|. Notice that 2 2 2 3 3n + 4 3 2(3n + 4) − 3(2n + n + 5) |− 7 − 3n| = − = = an − 2n2 + n + 5 2(2n2 + n+ 5) |2(2n2 + n+ 5)| 2 2 3n+ 7 3n+ 7n 10n 10 = < < = . 2(2n2 + n+ 5) 4n2 + 2n + 5 4n2 4n ≤ 10 10 Observe that if n ≥ N, then 10 4N . To fnd the condition on N we solve the inequality 4 4n N < ε 10 obtaining N > 4ε . The last step is to write a formal proof. Let ε > 0 and choose an integer N 10 satisfying N > 4ε . Such an N exists by the Archimedean property (Theorem 1.6.1). For every n ≥ N, one has 10 10 10 |an − a|≤ ≤ < 410 = ε. 4n 4N 4ε Therefore, 3n2 + 4 3 lim = . n→∞ 2n2 + n+ 5 2 ■ Example 2.1.4 Let {an} be given by 4n2 − 1 an = . 3n2 − n We claim limn →∞ an = 4/3. Let ε > 0. We search for a suitable N. First notice that 4n2 − 1 4 12n2 − 3 − 12n2 + 4n |4n − 3| |4n − 3| − = = = . 3n2 − n 3(3n2 − n) |3(3n2 − n)| 3|3n2 − n| 3 Since n ≥ 1, we have 4n > 3 and n2 ≥ n. Thus, |4n− 3| = 4n − 3 and |3n2 − n| = 3n2 − n. Also 4n− 3 < 4n and 3n − 1 ≥ 3n− n. Therefore, |4n− 3| 4n − 3 4n− 3 4n − 3 4n 4 = = ≤ < = . 3|3n2 − n| 3(3n2 − n) 3n(3n − 1) 3n(3n − n) 6n2 6n 4 Hence, if N > 6ε , we have, for n ≥ N 2 − 4 1 n 4 4 4 − < ≤ < ε. 3n2 − n 3 6n 6N Therefore, 2 − 4 1 n 4 lim = . n→∞ 3n2 − n 3
RkJQdWJsaXNoZXIy NTc4NTAz