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 limits
12 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
1. 1. LIMIT OF SEQUENCE nta 4.m+(-1)a n+(-1)2b 1)b n+bsin n 1 (-1)"(an+b) √n2+an+b n2+c(-1)n+1n+d 3. Vn+c+d an+ bsin n n2+an+b 1)ny2⊥d cn+dsin n Exercise 1.1. 15. Find the limits, p>0 sin 2.sin(n+1) asin n+b 3. mp nP+(-1 on np+c Example 1. 1.5. For a>0, the sequence vn+a-vn satisfies a 0<√n+a (n+a-vn)(n+a+n) n+a+vn n+a+√nn By mn* v分0 and the sandwich rule, we get limn→(Vm+a-√n)=0. Similar Argument also shows the limit for a <0 Example 1.1.6. The sequence Vn- satisfies 1<,n+2n+2 1+2 n By limn=oo(1+2n)=1+2. 0=1 and the sandwich rule, we get limn+ooV n= Exercise 1. 16 Show that limn-+oo(Vn+a-vn)=0 for a<0 Exercise 1.1.17. Use the idea of Example 1.1.5 to estimate 1 and then find n+2 limn→ Exercise 1.1.18. Show that limn-ooV n+6 1. You may need separate argument for a>b and a b Exercise 1.1.19. Find the limits 1.√m+a-vn+b /nta n+c+√n+
1.1. LIMIT OF SEQUENCE 13 1. √ n + a n + (−1)nb . 2. 1 √3 n2 + an + b . 3. √ n + c + d √3 n2 + an + b . 4. n + (−1)na n + (−1)nb . 5. (−1)n (an + b) n2 + c(−1)n+1n + d . 6. (−1)n (an + b) 2 + c (−1)nn2 + d . 7. cos √ n + a n + b sin n . 8. cos √ n + a √ n + b sin n . 9. an + b sin n cn + d sin n . Exercise 1.1.15. Find the limits, p > 0. 1. sin √ n np . 2. sin(n + 1) np + (−1)n . 3. a sin n + b np + c . 4. a cos(sin n) np − b sin n . Example 1.1.5. For a > 0, the sequence √ n + a − √ n satisfies 0 < √ n + a − √ n = ( √ n + a − √ n)(√ n + a + √ n) √ n + a + √ n = a √ n + a + √ n < a √ n . By limn→∞ a √ n = 0 and the sandwich rule, we get limn→∞( √ n + a − √ n) = 0. Similar argument also shows the limit for a < 0. Example 1.1.6. The sequence r n + 2 n satisfies 1 < r n + 2 n < n + 2 n = 1 + 2 1 n . By limn→∞ 1 + 2 1 n = 1 + 2 · 0 = 1 and the sandwich rule, we get limn→∞ r n + 2 n = 1. Exercise 1.1.16. Show that limn→∞( √ n + a − √ n) = 0 for a < 0. Exercise 1.1.17. Use the idea of Example 1.1.5 to estimate r n + 2 n − 1 and then find limn→∞ r n + 2 n . Exercise 1.1.18. Show that limn→∞ r n + a n + b = 1. You may need separate argument for a > b and a < b. Exercise 1.1.19. Find the limits. 1. √ n + a − √ n + b. 2. √ n + a √ n + c + √ n + d
14 CHAPTER 1. LIMIT n+c+Vn+d n2+bn+c 4. Vn+avn+b 11.√n2+an+b-√n2+cn+d 12. n+avn+b-vn+cVn+d 6.vn(vn+a-√n+b) n2+n+ 7.√n+c(√m+a-n+b) n+a 8.Vn+a+n+b-2√n+c n2+bn +c 9. an+b n2+n+1 n2+cn +d Exercise 11.20. Find the limits 1.√n+ asm n-√h+ b cos n. 5.√m+(-1)"(√m+a-√h+b) n+ sinn Vn+bcos n 6.√m2+an+sinn +bn+cos n 3.,/m+(-1)ya an+sinn n+(-1)mb n2+bn+cosn 4. Vn+a+sinn n2+an+b mn+c+(-1)n n+(-1 Exercise 11.21. Find the limits 3.n2(3n+a-n+b a 4. y/n(vn+a-vvn+b) Exercise 11.22. find the limits 5.4 1./n 2. n+1 Exercise 1.1. 23. Find the limits /n+ bcos 2n 2.(m2+(-1)nc 3.(m2+bm+c Exercise 1. 1.24. Suppose limn-oo n 1. Use the arithmetic rule and the sandwich rule to prove that, if In< 1, then limn-oo zn=l. Of course we expect the condition In<1 to be unnecessary. See Example 1.1.21
14 CHAPTER 1. LIMIT 3. √ n + a + √ n + b √ n + c + √ n + d . 4. √ n + a √ n + b √ n + c √ n + d . 5. √ n + a + b √ n + c + d . 6. √ n( √ n + a − √ n + b). 7. √ n + c( √ n + a − √ n + b). 8. √ n + a + √ n + b − 2 √ n + c. 9. r n n2 + n + 1 . 10. r n + a n2 + bn + c . 11. √ n2 + an + b − √ n2 + cn + d. 12. √ n + a √ n + b − √ n + c √ n + d. 13. n √ n2 + n + 1 . 14. n + a √ n2 + bn + c . 15. r n 2 + an + b n2 + cn + d . Exercise 1.1.20. Find the limits. 1. √ n + a sin n − √ n + b cos n. 2. r n + a sin n n + b cos n . 3. s n + (−1)na n + (−1)nb . 4. √ n + a + sin n √ n + c + (−1)n . 5. p n + (−1)n( √ n + a − √ n + b). 6. √ n2 + an + sin n − √ n2 + bn + cos n. 7. r n 2 + an + sin n n2 + bn + cos n . 8. √ n2 + an + b n + (−1)nc . Exercise 1.1.21. Find the limits. 1. √3 n + a − √3 n + b. 2. 3 r n + a n + b . 3. √3 n2( √3 n + a − √3 n + b). 4. √3 n( p3 √ n + a − p3 √ n + b). Exercise 1.1.22. Find the limits. 1. n − 2 n + 15 . 2. n − 2 n + 15.4 . 3. n − 2 n + 1− √ 2 . 4. n + a n + b p . Exercise 1.1.23. Find the limits. 1. √ n + a sin n √ n + b cos 2n p . 2. n 2 + an + b n2 + (−1)nc p . 3. n + a n2 + bn + c p . Exercise 1.1.24. Suppose limn→∞ xn = 1. Use the arithmetic rule and the sandwich rule to prove that, if xn ≤ 1, then limn→∞ x p n = 1. Of course we expect the condition xn ≤ 1 to be unnecessary. See Example 1.1.21
1. 1. LIMIT OF SEQUENCE 1.1.3 Some basic limits Using limn-oo -=0 and the sandwich rule, we may establish some basic limit Example 1.1.7. We show that lin 1. for a>0 First assume a≥1. Then In=ya-1≥0,and a=(1+xn)=1+nrn+ n(n-1 xn+…+xn>nrn 0≤n< By the sandwich rule and limn- 0, we get lim -oo n =0. Then by the arithmetic rule, this further implies lim va= lim(an+1)=lin 1=1 For the case0<a<I, let 1 >I. Then by the arithmetic rule lim v lim Example 1.1.8. Example 1.1.7 can be extended to Let n= Vn-1. Then we have In>0 for sufficiently large n(in fact, n 2 2 is enough), and 7(n-1) = Vn-T0(see Example 1. 1. 1 or 1.1.3)and the sandwich rule, we get mn-oo In=0. This further im lim v n→0 n→△o
1.1. LIMIT OF SEQUENCE 15 1.1.3 Some Basic Limits Using limn→∞ 1 n = 0 and the sandwich rule, we may establish some basic limits. Example 1.1.7. We show that limn→∞ √n a = 1, for a > 0. First assume a ≥ 1. Then xn = √n a − 1 ≥ 0, and a = (1 + xn) n = 1 + nxn + n(n − 1) 2 x 2 n + · · · + x n n > nxn. This implies 0 ≤ xn < a n . By the sandwich rule and limn→∞ a n = 0, we get limn→∞ xn = 0. Then by the arithmetic rule, this further implies limn→∞ √n a = limn→∞ (xn + 1) = limn→∞ xn + 1 = 1. For the case 0 < a ≤ 1, let b = 1 a ≥ 1. Then by the arithmetic rule, limn→∞ √n a = limn→∞ 1 √n b = 1 limn→∞ √n b = 1 1 = 1. Example 1.1.8. Example 1.1.7 can be extended to limn→∞ √n n = 1. Let xn = √n n − 1. Then we have xn > 0 for sufficiently large n (in fact, n ≥ 2 is enough), and n = (1 + xn) n = 1 + nxn + n(n − 1) 2 x 2 n + · · · + x n n > n(n − 1) 2 x 2 n . This implies 0 ≤ xn < √ 2 √ n − 1 . By limn→∞ √ 2 √ n − 1 = 0 (see Example 1.1.1 or 1.1.3) and the sandwich rule, we get limn→∞ xn = 0. This further implies limn→∞ √n n = limn→∞ xn + 1 = 1
16 CHAPTER 1. LIMIT Example 1. 1.9. The following "n-th root type"limits can be compared with the limits in Examples 1.1.7 and 1. 1.8 1<wn+1<V2 2n 1<nn+i vn 1 (vm)4 By Examples 1.1.7, 1.1.8 and the arithmetic rule, the sequences on the right converge to 1. Then by the sandwich rule, we get lim Vn+1-=lim nntr=lim(n2-n n→0 n→∞0 Example 1.1.10. We have 3=3n<n+3n<√3n+3n=3v2 By Example 1. 1.7, we have limn-oo 3V2=3. Then by the sandwich rule, we get limn_o v2n+3m=3 For another example, we have n-1+2.3 Taking the n-th root, we get y3n-2n>3 1 v llm 3 and the sandwich rule, we get limn-oo V3n-2n=3 Exercise 1. 1.25. Prove that if a< an< b for some constants a, b>0 and sufficiently big n, then limn→xn=1. Exercise 1.1.26. Find the limits. a>0 5.(an+b) 9.(an+b) 6.(an2+b)后 (an+b) 3. 4.(7+1)元 12.(an2+b)n+cm+了 Exercise 1.1.27. Find the limits. a>0
16 CHAPTER 1. LIMIT Example 1.1.9. The following “n-th root type” limits can be compared with the limits in Examples 1.1.7 and 1.1.8 1 < √n n + 1 < √n 2n = √n 2 √n n, 1 < n 1 n+1 < √n n, 1 < (n 2 − n) n n2−1 < (n 2 ) n n2/2 = (√n n) 4 . By Examples 1.1.7, 1.1.8 and the arithmetic rule, the sequences on the right converge to 1. Then by the sandwich rule, we get limn→∞ √n n + 1 = limn→∞ n 1 n+1 = limn→∞ (n 2 − n) n n2−1 = 1. Example 1.1.10. We have 3 = √n 3 n < √n 2 n + 3n < √n 3 n + 3n = 3√n 2. By Example 1.1.7, we have limn→∞ 3 √n 2 = 3. Then by the sandwich rule, we get limn→∞ √n 2 n + 3n = 3. For another example, we have 3 n > 3 n − 2 n = 3n−1 + 2 · 3 n−1 − 2 · 2 n−1 > 3 n−1 . Taking the n-th root, we get 3 > √n 3 n − 2 n > 3 1 √n 3 . By limn→∞ 3 1 √n 3 = 3 and the sandwich rule, we get limn→∞ √n 3 n − 2 n = 3. Exercise 1.1.25. Prove that if a ≤ xn ≤ b for some constants a, b > 0 and sufficiently big n, then limn→∞ √n xn = 1. Exercise 1.1.26. Find the limits, a > 0. 1. n 1 2n . 2. n 2 n . 3. n c n . 4. (n + 1) c n . 5. (an + b) c n . 6. (an2 + b) c n . 7. ( √ n + 1) c n . 8. (n − 2) 1 n+3 . 9. (an + b) c n+d . 10. (an + b) cn n2+dn+e . 11. (an2 + b) c n+d . 12. (an2 + b) cn+d n2+en+f . Exercise 1.1.27. Find the limits, a > 0