35 ■ Example 2.1.3 Consider the sequence {an}where an = 3n2 +4 2n2 +n+5 . We will prove directly from the definition that this sequence converges toa=3/2. Let ε >0. We first search for a suitable positive integer N. To that end, we simplify and estimate the expression|an −a|. Notice that an − 3 2 = 3n2 +4 2n2 +n+5− 3 2 = 2(3n2 +4)−3(2n2 +n+5) 2(2n2 +n+5) = | − 7−3n| |2(2n2 +n+5)| = 3n+7 2(2n2 +n+5) < 3n+7n 4n2 +2n+5 < 10n 4n2 = 10 4n . Observe that if n≥N, then 10 4n ≤ 10 4N. To find the condition on Nwe solve the inequality 10 4N <ε obtaining N>10 4ε . The last step is to write a formal proof. Let ε >0 and choose an integer N satisfyingN>10 4ε . Such anNexists by the Archimedean property (Theorem 1.6.1). For everyn≥N, one has |an −a| ≤ 10 4n ≤ 10 4N < 10 410 4ε =ε. Therefore, lim n→∞ 3n2 +4 2n2 +n+5 = 3 2 . ■ Example 2.1.4 Let {an}be given by lim n→∞ 4n2 −1 3n2 −n . We claim limn→∞an =4/3. Let ε >0. We search for a suitable N. First notice that 4n2 −1 3n2 −n− 4 3 = 12n2 −3−12n2 +4n 3(3n2 −n) = | 4n−3| |3(3n2 −n)| = | 4n−3| 3|3n2 −n| . Since n≥1, we have 4n>3 andn2 ≥n. Thus, |4n−3| =4n−3 and|3n2 −n| =3n2 −n. Also 4n−3<4nand 3n−1≥3n−n. Therefore, |4n−3| 3|3n2 −n| = 4n−3 3(3n2 −n) = 4n−3 3n(3n−1) ≤ 4n−3 3n(3n−n) < 4n 6n2 = 4 6n . Hence, if N> 4 6ε , we have, for n≥N 4n2 −1 3n2 −n− 4 3 < 4 6n ≤ 4 6N <ε. Therefore, lim n→∞ 4n2 −1 3n2 −n = 4 3 .
RkJQdWJsaXNoZXIy NTc4NTAz