1. 1. LIMIT OF SEQUENCE 1. Vn+sinn 3. Vn+(1)"sinn 5.(n-cos n)n+ 2. van +bsin n 4.van+(1)mosin 6.(an+bsin n)n2+ccos r Exercise 1. 28. Find the limits, p, q>0 1. V/nP +sin n 3.n+m 4. n-v/mp +n? Exercise 11 29. Find the limits 1.5n-4 4.v54-3·4-2n 2.5n-3·4 5.v42m-1-5n 8.(52-4) 3.v5n-3·4n+2n 6.V42m-1+(-1)5 9.(5-4) Exercise 1.1.30. Find the limits. a>b>0 1. van+bn n h2n+1 7.(an+b)n- 2. van-bn 5. van tbn 3.van+(-1)b 6. nvan-bn (-1)2b2)n2-i Exercise 1.1.31. For a, b, c>0, find limn_oo van+bn +cn Exercise 1. 1.32. For a 1, prove limn-oo va=l by using (Va-1)((va)n-+(va) at Example 1.1.11. We show that First assume 0< a< l and write a=1+6 Then b>0 and (+b=,m(n-1)2<mb 2 m If-1<36=0 and the sandwich rule, we get limn-ooQ=0 By limn→ a <0, then 0< a I and limn-oo a"= limn-oo a n=0. By Exercise 1, we get limn→oal xample 1.1.12. Ex 1.1.11 can be extended t 0, for a<1
1.1. LIMIT OF SEQUENCE 17 1. √n n + sin n. 2. √n an + b sin n. 3. pn n + (−1)n sin n. 4. pn an + (−1)nb sin n. 5. (n − cos n) 1 n+sin n . 6. (an + b sin n) n n2+c cos n . Exercise 1.1.28. Find the limits, p, q > 0. 1. √n np + sin n. 2. √n np + nq . 3. n+2√ np + nq . 4. n−√2 np + nq . Exercise 1.1.29. Find the limits. 1. √n 5 n − 4 n. 2. √n 5 n − 3 · 4 n. 3. √n 5 n − 3 · 4 n + 2n. 4. √n 5 n − 3 · 4 n − 2 n. 5. √n 4 2n−1 − 5 n. 6. pn 4 2n−1 + (−1)n5 n. 7. (5n − 4 n ) 1 n+1 . 8. (5n − 4 n ) 1 n−2 . 9. (5n − 4 n ) n+1 n2+1 . Exercise 1.1.30. Find the limits, a > b > 0. 1. √n a n + b n. 2. √n a n − b n. 3. pn a n + (−1)nb n. 4. √n a nb 2n+1. 5. n+2√ a n + b n. 6. n−√2 a n − b n. 7. (a n + b n ) n n2−1 . 8. (a n − b n ) n n2−1 . 9. (a n − (−1)n b n ) n n2−1 . Exercise 1.1.31. For a, b, c > 0, find limn→∞ √n a n + b n + c n. Exercise 1.1.32. For a ≥ 1, prove limn→∞ √n a = 1 by using a − 1 = ( √n a − 1) ( √n a) n−1 + ( √n a) n−2 + · · · + √n a + 1 . Example 1.1.11. We show that limn→∞ a n = 0, for |a| < 1. First assume 0 < a < 1 and write a = 1 1 + b . Then b > 0 and 0 < an = 1 (1 + b) n = 1 1 + nb + n(n − 1) 2 b 2 + · · · + b n < 1 nb . By limn→∞ 1 nb = 0 and the sandwich rule, we get limn→∞ a n = 0. If −1 < a < 0, then 0 < |a| < 1 and limn→∞ |a n | = limn→∞ |a| n = 0. By Exercise 1.1.11, we get limn→∞ a n = 0. Example 1.1.12. Example 1.1.11 can be extended to limn→∞ nan = 0, for |a| < 1
18 CHAPTER 1. LIMIT This follows from (1+b) n(n-1 (n-1)b2 the limit limn+∞(n-1)2 =0 and the sandwich rule Exercise 1. 1.58 gives further extension of the limit ple1.1.13 lim 4. For n>4 4n4.4·4·44 44·4·4.444°1 0 451 Exercise 1. 1.59 suggests how to show the limit in gener/=0 By lim =0 and the sandwich rule. we get li 13mx1.1 Tohe secondis wo ways, im时b la< Exercise 1. 1.34. Show that limn-yoo ns4an"=0 for la< 1. What about limn-oo n-34an? What about limn_oo nPa"? Exercise 1.135. Show that limn-oo n=0 for a=5.4 and a=-54 Exercise 1.136. Show that limno vn! =0 for a= 5.4 and a=-54 Exercise 1.1.37. Show that limn-oo (2n)! =0 for any a Exercise 1.1.38. Find the limits n+1 3.m990.99a n+22 (1+2n)2 1.001 4n Exercise 1.1.39. Find the limits. Some convergence depends on a and p. You may try some special values of a and p first
18 CHAPTER 1. LIMIT This follows from 0 < nan = n (1 + b) n = n 1 + nb + n(n − 1) 2 b 2 + · · · + b n < n n(n − 1) 2 b 2 = 2 (n − 1)b 2 , the limit limn→∞ 2 (n − 1)b 2 = 0 and the sandwich rule. Exercise 1.1.58 gives further extension of the limit. Example 1.1.13. We show that limn→∞ a n n! = 0, for any a for the special case a = 4. For n > 4, we have 0 < 4 n n! = 4 · 4 · 4 · 4 1 · 2 · 3 · 4 · 4 5 · 4 6 · · · 4 n ≤ 4 · 4 · 4 · 4 1 · 2 · 3 · 4 · 4 n = 4 5 4! 1 n . By limn→∞ 4 5 4! 1 n = 0 and the sandwich rule, we get limn→∞ 4 n n! = 0. Exercise 1.1.59 suggests how to show the limit in general. Exercise 1.1.33. Show that limn→∞ n 2a n = 0 for |a| < 1 in two ways. The first is by using the ideas from Examples 1.1.11 and 1.1.12. The second is by using limn→∞ nan = 0 for |a| < 1. Exercise 1.1.34. Show that limn→∞ n 5.4a n = 0 for |a| < 1. What about limn→∞ n −5.4a n ? What about limn→∞ n pa n ? Exercise 1.1.35. Show that limn→∞ a n n! = 0 for a = 5.4 and a = −5.4. Exercise 1.1.36. Show that limn→∞ a n √ n! = 0 for a = 5.4 and a = −5.4. Exercise 1.1.37. Show that limn→∞ n!a n (2n)! = 0 for any a. Exercise 1.1.38. Find the limits. 1. n + 1 2 n . 2. n 2 2 n . 3. n 990.99n . 4. (n 2 + 1)1001 1.001n−2 . 5. n + 2n 3 n . 6. n2 n + (−3)n 4 n . 7. n3 n (1 + 2n) 2 . 8. 5 n − n6 n+1 3 2n−1 − 2 3n+1 . Exercise 1.1.39. Find the limits. Some convergence depends on a and p. You may try some special values of a and p first
1. 1. LIMIT OF SEQUENCE 19 4. 7 10. W/n! Vni, 1.(2n) nInP (2n)! Exercise 1.1.40. Find the limits 3n+5 n2+n3+5! +nl!+(n-1)! n2+3n+5 Exercise 1. 1.41. Prove Vn!> Then use this to prove limn→∞ /n! Exercise 1.1.42. Prove (2n)! n+1 or n>1. Then use this to prove limn-too mn-n-+0(2n/)/0. What about lim (n!)-? 1.1. 4 Order rule The following property reflects the intuition that bigger sequence should have bigger Proposition1.1.5( Order rule). Suppose limn→∞oxrn= l and limn→∞yhn=k 1.In≤ yn for sufficiently big n, then l≤k 2. If l< k, then n yn for sufficiently big n By taking yn =l, we get the following special cases of the property for a con- verging sequence I'n 1.Ifxn≤ l for sufficiently big n, then lim→sxn≤l 2. If limn-yoo In < L, then an I for sufficiently big n Similar statements with reversed inequalities also hold(see Exercise 1. 1.43) Note the non-strict inequality in the first statement of Proposition 1. 1.5 and the strict ality the second statement. Fo we have r n but lim→ n lim→xyn, The example also satisfies lim→sxn≥limn→sn but In y yn, even for sufficiently big n
1.1. LIMIT OF SEQUENCE 19 1. n pa n . 2. a n np . 3. n p a n . 4. 1 npa n . 5. n p n! . 6. n pa n n! . 7. n p n!a n . 8. n p √ n! . 9. n pa n √ n! . 10. n pa n √3 n! . 11. n!a n (2n)!. 12. n!n pa n (2n)! . Exercise 1.1.40. Find the limits. 1. n 2 + 3n + 5n n! . 2. n 2 + 3n + 5n n! − n2 + 2n . 3. n 2 + n3 n + 5! n! . 4. n 23 n+5 + 5 · (n − 1)! (n + 1)! . 5. n 2 + n! + (n − 1)! 3 n − n! + (n − 1)!. 6. 2 nn! + 3n (n − 1)! 4 n(2n − 1)! − 5 nn! . Exercise 1.1.41. Prove √n n! > rn 2 . Then use this to prove limn→∞ 1 √n n! = 0. Exercise 1.1.42. Prove n! nn < 1 n and (n!)2 (2n)! < 1 n + 1 for n > 1. Then use this to prove limn→∞ n! nn = limn→∞ (n!)2 (2n)! = 0. What about limn→∞ (n!)k (kn)!? 1.1.4 Order Rule The following property reflects the intuition that bigger sequence should have bigger limit. Proposition 1.1.5 (Order Rule). Suppose limn→∞ xn = l and limn→∞ yn = k. 1. If xn ≤ yn for sufficiently big n, then l ≤ k. 2. If l < k, then xn < yn for sufficiently big n. By taking yn = l, we get the following special cases of the property for a converging sequence xn. 1. If xn ≤ l for sufficiently big n, then limn→∞ xn ≤ l. 2. If limn→∞ xn < l, then xn < l for sufficiently big n. Similar statements with reversed inequalities also hold (see Exercise 1.1.43). Note the non-strict inequality in the first statement of Proposition 1.1.5 and the strict inequality the second statement. For example, we have xn = 1 n2 < yn = 1 n , but limn→∞ xn 6< limn→∞ yn. The example also satisfies limn→∞ xn ≥ limn→∞ yn but xn 6≥ yn, even for sufficiently big n
CHAPTER 1. LIMIT Exercise 1. 1.43. Explain how to get the following special cases of the order rule 1.Ifrn≥ I for sufficiently big n, then limn→rn≥l 2. If limn-*oo In>l, then n >l for sufficiently big n. 2n2+n Example 1.1.14. By lim =2 and the order rule. we know 1< n+ 2n2+n 2m2+n m2-n+i <3 for sufficiently big n. This implies 1 sufficiently big n. By limn-oo V3= 1 and the sandwich rule, we get 2 lim Example 1.1.15. We showed limn-oo V3n-2n=3 in Example 1.1.10. Here we use a different method, with the help of the order rule Byv3-2=311 nly need to find lin (3) 1.11, we have limn→∞ 1. By the order rule, therefore we have 1 1 2 or sufficiently big n. This implies that for sufficiently big n. Then by Example 1.1.7 and the sandwich rule, we get limn→+∞ 1. and we conclude that lm2=3lmn/1-(2)=3 Exercise 1.1.44. Prove that if limn-oo n=l>0, then limn-oo Vn=l. Moreover, find a sequence satisfying limn-yoo n=0 and limn-yoo vn=l. Can we have an converging to 0 and v/En converging to 0. 32? Exercise 1.1.45. Find the limits
20 CHAPTER 1. LIMIT Exercise 1.1.43. Explain how to get the following special cases of the order rule. 1. If xn ≥ l for sufficiently big n, then limn→∞ xn ≥ l. 2. If limn→∞ xn > l, then xn > l for sufficiently big n. Example 1.1.14. By limn→∞ 2n 2 + n n2 − n + 1 = 2 and the order rule, we know 1 < 2n 2 + n n2 − n + 1 < 3 for sufficiently big n. This implies 1 < n r 2n 2 + n n2 − n + 1 < √n 3 for sufficiently big n. By limn→∞ √n 3 = 1 and the sandwich rule, we get limn→∞ n s 2n 2 + n n2 − n + 1 = 1. Example 1.1.15. We showed limn→∞ √n 3 n − 2 n = 3 in Example 1.1.10. Here we use a different method, with the help of the order rule. By √n 3 n − 2 n = 3 n s 1 − 2 3 n , we only need to find limn→∞ n s 1 − 2 3 n . By Example 1.1.11, we have limn→∞ 1 − 2 3 n = 1. By the order rule, therefore, we have 1 2 < 1 − 2 3 n < 2 for sufficiently big n. This implies that 1 √n 2 < n s 1 − 2 3 n < √n 2 for sufficiently big n. Then by Example 1.1.7 and the sandwich rule, we get limn→∞ n s 1 − 2 3 n = 1, and we conclude that limn→∞ √n 3 n − 2 n = 3 limn→∞ n s 1 − 2 3 n = 3. Exercise 1.1.44. Prove that if limn→∞ xn = l > 0, then limn→∞ √n xn = 1. Moreover, find a sequence satisfying limn→∞ xn = 0 and limn→∞ √n xn = 1. Can we have xn converging to 0 and √n xn converging to 0.32? Exercise 1.1.45. Find the limits
1. 1. LIMIT OF SEQUENCE 4 /5n-n4n (-1)n (52-(-1)"n42) #1 Exercise 1146. find the limits Bn n 52-(-1)n4 man sin Exercise 1. 47. Find the limits. a>b /n+11 6. va+bn 10.(ax2+b) an+I+(1)mbn 7. van+bn 11.(n+1)an+b2)n2-i 3."van+1+(-1)b.8.nrmn+(n2+1) 4. V4an-5bn 12.(an+b2)m2-1 5.v4an+5b2n+1 9. 13.(an+(-1)2b2) G=1) Exercise 1.1. 48. Find the limits. a.b. c>0 1. Nian+nbn +2cn V(n+sinn)an+bn+ (n+b2(1+nc) Exercise 1.1.49. Suppose a polynomial p(n)=apnP+an-inP-+.+ain+ao has leading coefficient ap>0. Prove that p(n)>0 for sufficiently big n Exercise 1. 1.50. Suppose a, b, c>0, and p, g, r are polynomials with positive leading coef- ficients. Find the limit of vp(n)am+q(n)bn +r(n)en Exercise 1. 1.51. Find the limits, a, b, p, q>0 1. vanP+ bsin n. 3.n*vanp + bn? 5. VanP +bnq
1.1. LIMIT OF SEQUENCE 21 1. √n 5 n − n4 n. 2. n+2√ 5 n − n4 n. 3. n−√2 5 n − n4 n. 4. pn 5 n − (−1)nn4 n. 5. (5n − n4 n ) n−1 n2+1 . 6. (5n−(−1)nn4 n ) n+(−1)n n2+1 . Exercise 1.1.46. Find the limits. 1. n r 1 n 5 n − n4 n. 2. n r 1 n 5 n − (−1)nn4 n. 3. n+2r 1 n 5 n − n4 n sin n. 4. (n 24 2n−1 − 5 n ) n−1 n2+1 . 5. n r n2 n + 4n+1 + 3 n−1 n . 6. n−2 r n2 3n + 3 2n−1 n2 . 7. n−2 r 2 3n + n − 1 n2 + 1 3 2n−1. 8. n2 3n + 3 2n−1 n2 n−1 n2 . 9. n2 3n + 3 2n−1 n2 1 n2 . Exercise 1.1.47. Find the limits, a > b. 1. √n a n+1 + b n. 2. pn a n+1 + (−1)nb n. 3. n−p2 a n+1 + (−1)nb n. 4. √n 4a n − 5b n. 5. √n 4a n + 5b 2n+1. 6. √n a + b n. 7. √n an + b n. 8. n−p2 nan + (n2 + 1)b n. 9. n+2r 1 n a n + nbn+1. 10. (a n + b n ) n+1 n2+1 . 11. ((n + 1)a n + b n ) n n2−1 . 12. (a n + b n ) 1 n2−1 . 13. (a n + (−1)n b n ) (−1)n n2−1 . Exercise 1.1.48. Find the limits, a, b, c > 0. 1. √n n2a n + nbn + 2c n. 2. pn a n(b n + 1) + ncn. 3. pn (n + sin n)a n + b n + n2c n. 4. pn a n(n + b n(1 + ncn)). Exercise 1.1.49. Suppose a polynomial p(n) = apn p +an−1n p−1 +· · ·+a1n+a0 has leading coefficient ap > 0. Prove that p(n) > 0 for sufficiently big n. Exercise 1.1.50. Suppose a, b, c > 0, and p, q, r are polynomials with positive leading coef- ficients. Find the limit of pn p(n)a n + q(n)b n + r(n)c n. Exercise 1.1.51. Find the limits, a, b, p, q > 0. 1. √n anp + b sin n. 2. √n anp + bnq . 3. n+2√ anp + bnq . 4. n−√2 anp + bnq . 5. n√2 anp + bnq . 6. (anp + bnq ) 1 n2−1