12 CHAPTER 1. LIMIT By limn → =2v2 lim n 0(arithmetic rule used) and the sandwich rule, we get lim→a 0. n-1 Example 1. 1.4. By -1<sin n 1, we have In n nn-n’√n+1-√n+ SIn n By limn- -=0, -limn+=0 and the sandwich rule, we get limn+oo n0 Moreover, by lim→∞ 1=limn-o Vn-i=0(see argument in Example 1.1.1)and the sandwich rule, we get limn-yoo 0 √n+sin Exercise1.1.11. Prove that limn→xn=0 implies limn→∞xn=0. Exercise 1.1. 12. Find the limits, a>0 Exercise 1.1.13. Find the limits 14. c n-1 hn+(-1) + 11 sinn+(-1)2 √n+(-1)n n+(-1)n 3vm+2 (-1)2(n+10)2-103 1.1. 14. Find the limits12 CHAPTER 1. LIMIT By limn→∞ 2 √ 2 √ n = 2√ 2 limn→∞ 1 √ n = 0 (arithmetic rule used) and the sandwich rule, we get limn→∞ √ n + 1 n − 1 = 0. Example 1.1.4. By −1 ≤ sin n ≤ 1, we have − 1 n ≤ sin n n ≤ 1 n , 1 √ n + 1 ≤ 1 √ n + sin n ≤ 1 √ n − 1 . By limn→∞ 1 n = 0, − limn→∞ 1 n = 0 and the sandwich rule, we get limn→∞ sin n n = 0. Moreover, by limn→∞ 1 √ n + 1 = limn→∞ 1 √ n − 1 = 0 (see argument in Example 1.1.1) and the sandwich rule, we get limn→∞ 1 √ n + sin n = 0. Exercise 1.1.11. Prove that limn→∞ |xn| = 0 implies limn→∞ xn = 0. Exercise 1.1.12. Find the limits, a > 0. 1. 1 √ 3n − 4 . 2. √ 2n + 3 4n − 1 . 3. 1 √ an + b . 4. √ an + b cn + d . Exercise 1.1.13. Find the limits. 1. cos n n . 2. (−1)n n . 3. sin √ n n . 4. cos n √ n − 2 . 5. 1 n + (−1)n . 6. cos n n + (−1)n . 7. cos n p n + (−1)n2 . 8. cos n p n + sin √ n . 9. (−1)n p n + (−1)n . 10. 2 + (−1)n3 √3 n2 − 2 cos n . 11. sin n + (−1)n cos n √ n + (−1)n . 12. |sin n + cos n| n . 13. 3 √ n + 2 2n + (−1)n3 . 14. √ n sin n + cos n n − 1 . 15. n + sin √ n n + cos 2n . 16. √ n + sin n √ n − cos n . 17. (−1)n (n + 1) n2 + (−1)n+1 . 18. (−1)n (n + 10)2 − 1010 10(−1)nn2 − 5 . Exercise 1.1.14. Find the limits