Chapter 1 Limit 1.1 Limit of Sequence A sequence is an infinite list 1,2,…,n The n-th term of the sequence is an, and n is the index of the term. In this course. we will always assume that all the terms are real numbers. Here are some examples n yn=2:2,2,2 1 1 2’’n 1.-1.1 Un sin n in 1. sin 2. sin 3 Sin n Note that the index does not have to start from 1. For example, the sequence Un actually starts from n=0 (or any even integer). Moreover, a sequence does not have to be given by a formula. For example, the decimal expansions of T give a sequence n:3,3.1,3.14,3.141,3.1415,3.14159,3.141592 If n is the number of digits after the decimal point, then the sequence wn starts at n=0 Now we look at the trend of the examples above as n gets bigger. We find that n gets bigger and can become as big as we want. On the other hand, yn remains constant,zn gets smaller and can become as small as we want. This means that Un approaches 2 and zn approaches 0. Moreover, un and Un jump around and do not approach anything. Finally, wn is equal to T up to the n-th decimal place, and therefore approaches T
Chapter 1 Limit 1.1 Limit of Sequence A sequence is an infinite list x1, x2, . . . , xn, . . . . The n-th term of the sequence is xn, and n is the index of the term. In this course, we will always assume that all the terms are real numbers. Here are some examples xn = n: 1, 2, 3, . . . , n, . . . ; yn = 2: 2, 2, 2, . . . , 2, . . . ; zn = 1 n : 1, 1 2 , . . . , 1 n , . . . ; un = (−1)n : 1, −1, 1, . . . , (−1)n , . . . ; vn = sin n: sin 1, sin 2, sin 3, . . . , sin n, . . . . Note that the index does not have to start from 1. For example, the sequence vn actually starts from n = 0 (or any even integer). Moreover, a sequence does not have to be given by a formula. For example, the decimal expansions of π give a sequence wn : 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, . . . . If n is the number of digits after the decimal point, then the sequence wn starts at n = 0. Now we look at the trend of the examples above as n gets bigger. We find that xn gets bigger and can become as big as we want. On the other hand, yn remains constant, zn gets smaller and can become as small as we want. This means that yn approaches 2 and zn approaches 0. Moreover, un and vn jump around and do not approach anything. Finally, wn is equal to π up to the n-th decimal place, and therefore approaches π. 7
8 CHAPTER 1. LIMIT 日 日日 Figure 1. 1.1: Sequ Definition 1.1.1 (Intuitive). If In approaches a finite number I when n gets bigge and bigger, then we say that the sequence n converges to the limit l and write lim A sequence diverges if it does not approach a specific finite number when n gets The sequences n, Zn, Un converge respectively to 2, 0 and T. The sequences n, un, Un diverge. Since the limit describes the behavior when n gets very big, we have the following propert Proposition 1. 2. If yn is obtained from Tn by adding, deleting, or changing finitely many terms, then lim→oxn=limn→yn The equality in the proposition means that In converges if and only if yn con verges. Moreover, the two limits have equal value when both converge ample 1.. The sequence Vn+2 is obtained from Vn by deleting the first two terms. By limn→sJn =0 and Proposition 1. 1.2, we get lim, n+2 In general, we have limn-oo n+k= limn-yoo n for any integer k The example assumes limn-+oo 0, which is supposed to be intuitively obv ous. Although mathematics is inspired by intuition, a critical feature of mathematics is rigorous logic. This means that we need to be clear what basic facts are assumed in any argument. For the moment, we will always assume that we already know
8 CHAPTER 1. LIMIT n xn yn zn un vn wn Figure 1.1.1: Sequences. Definition 1.1.1 (Intuitive). If xn approaches a finite number l when n gets bigger and bigger, then we say that the sequence xn converges to the limit l and write limn→∞ xn = l. A sequence diverges if it does not approach a specific finite number when n gets bigger. The sequences yn, zn, wn converge respectively to 2, 0 and π. The sequences xn, un, vn diverge. Since the limit describes the behavior when n gets very big, we have the following property. Proposition 1.1.2. If yn is obtained from xn by adding, deleting, or changing finitely many terms, then limn→∞ xn = limn→∞ yn. The equality in the proposition means that xn converges if and only if yn converges. Moreover, the two limits have equal value when both converge. Example 1.1.1. The sequence 1 √ n + 2 is obtained from 1 √ n by deleting the first two terms. By limn→∞ 1 √ n = 0 and Proposition 1.1.2, we get limn→∞ 1 √ n = limn→∞ 1 √ n + 2 = 0. In general, we have limn→∞ xn+k = limn→∞ xn for any integer k. The example assumes limn→∞ 1 √ n = 0, which is supposed to be intuitively obvious. Although mathematics is inspired by intuition, a critical feature of mathematics is rigorous logic. This means that we need to be clear what basic facts are assumed in any argument. For the moment, we will always assume that we already know
1. 1. LIMIT OF SEQUENCE imn-ooC= c and limn-oo=0 for p>0. After the two limits are rigorously established in Examples 1.2.2 and 1.2.3, the conclusions based on the two limits become solid 1.1.1 Arithmetic rule Intuitively, if a is close to 3 and y is close to 5, then the arithmetic combinations t+y and ry are close to 3+5=8 and 3.5= 15. The intuition leads to the following property of limit Proposition1.13( Arithmetic rule). Suppose limn→xn= l and limn→∞n=k lim (an +yn)=l+k, lim can=c, lim Ingn=kl, lim En I n→0 where c is a constant and kf0 in the last equality The proposition says limn→(xn+mn)=limn→xn+limn→n. However,the equality is of different nature from the equality in Proposition 1.1.2, because the convergence of the limits on two sides are not equivalent: If the two limits on the ght converge, then the limit on the left also converges and the two sides are equal However, for In=(1)" and n =(1)n+, the limit limn-yoo(an +yn)=0 on the left converges. but both limits on the right diverge Exercise111. Explain that limn→∞n= l if and only if limn→∞(xn-l)=0. Exercise 1. 1.2. Suppose In and yn converge. Explain that limn-yoo nin=0 implies either limn-oo In=0 or limn-+oo n =0. Moreover, explain that the conclusion fails if an and n are not assumed to converge Example 1. 1.2. We have 2+ 2m2+n lim li 2+lin mn→1-limn→∞=+limn+nmn→ lin 2+0 0+0·0
1.1. LIMIT OF SEQUENCE 9 limn→∞ c = c and limn→∞ 1 np = 0 for p > 0. After the two limits are rigorously established in Examples 1.2.2 and 1.2.3, the conclusions based on the two limits become solid. 1.1.1 Arithmetic Rule Intuitively, if x is close to 3 and y is close to 5, then the arithmetic combinations x+y and xy are close to 3+5 = 8 and 3·5 = 15. The intuition leads to the following property of limit. Proposition 1.1.3 (Arithmetic Rule). Suppose limn→∞ xn = l and limn→∞ yn = k. Then limn→∞ (xn + yn) = l + k, limn→∞ cxn = cl, limn→∞ xnyn = kl, limn→∞ xn yn = l k , where c is a constant and k 6= 0 in the last equality. The proposition says limn→∞(xn + yn) = limn→∞ xn + limn→∞ yn. However, the equality is of different nature from the equality in Proposition 1.1.2, because the convergence of the limits on two sides are not equivalent: If the two limits on the right converge, then the limit on the left also converges and the two sides are equal. However, for xn = (−1)n and yn = (−1)n+1, the limit limn→∞(xn + yn) = 0 on the left converges, but both limits on the right diverge. Exercise 1.1.1. Explain that limn→∞ xn = l if and only if limn→∞(xn − l) = 0. Exercise 1.1.2. Suppose xn and yn converge. Explain that limn→∞ xnyn = 0 implies either limn→∞ xn = 0 or limn→∞ yn = 0. Moreover, explain that the conclusion fails if xn and yn are not assumed to converge. Example 1.1.2. We have limn→∞ 2n 2 + n n2 − n + 1 = limn→∞ 2 + 1 n 1 − 1 n + 1 n2 = limn→∞ 2 + 1 n limn→∞ 1 − 1 n + 1 n2 = limn→∞ 2 + limn→∞ 1 n limn→∞ 1 − limn→∞ 1 n + limn→∞ 1 n · limn→∞ 1 n = 2 + 0 1 − 0 + 0 · 0 = 2
10 CHAPTER 1. LIMIT The arithmetic rule is used in the second and third equalities. The limits limn-ooc 1 c and lim =0 are used in the fourth equalit Exercise 1.1.3. Find the limits 留++ (n+1)(n+2 2n3+3n 6 n2+3)3 (n+1)(n+2) Exercise 1.1.4. Find the limits ++ (√n+1)(√m+2) 3=切 2n-1 4n+1 (n+1)(n+ 9.( Exercise 1.1.5. Find the limits a +6 n+bvn+c (c√m+d) n (c√+d)3 n+a cn+d (a√n+b)2 m2+bn+c (√m+a)(√m+b) (cVn+d) Exercise 1.1.6. Show that 0. if lim …+a1n+a0 +bin+bo ifp= g and b≠0. bq Exercise 1.1.7. Find the limits 20 lo (2n+1)2-1 10m2-5 10n-5 Exercise 1.1. 8. Find the limits
10 CHAPTER 1. LIMIT The arithmetic rule is used in the second and third equalities. The limits limn→∞ c = c and limn→∞ 1 n = 0 are used in the fourth equality. Exercise 1.1.3. Find the limits. 1. n + 2 n − 3 . 2. n + 2 n2 − 3 . 3. 2n 2 − 3n + 2 3n2 − 4n + 1 . 4. n 3 + 4n 2 − 2 2n3 − n + 3 . 5. (n + 1)(n + 2) 2n2 − 1 . 6. 2n 2 − 1 (n + 1)(n + 2). 7. (n 2 + 1)(n + 2) (n + 1)(n2 + 2). 8. (2 − n) 3 2n3 + 3n − 1 . 9. (n 2 + 3)3 (n3 − 2)2 . Exercise 1.1.4. Find the limits. 1. √ n + 2 √ n − 3 . 2. √ n + 2 n − 3 . 3. 2 √ n − 3n + 2 3 √ n − 4n + 1 . 4. √3 n + 4√ n − 2 2 √3 n − n + 3 . 5. ( √ n + 1)(√ n + 2) 2n − 1 . 6. 2n − 1 ( √ n + 1)(√ n + 2). 7. ( √ n + 1)(n + 2) (n + 1)(√ n + 2). 8. (2 − √3 n) 3 2 √3 n + 3n − 1 . 9. ( √3 n + 3)3 ( √ n − 2)2 . Exercise 1.1.5. Find the limits. 1. n + a n + b . 2. √ n + a n + b . 3. n + a n2 + bn + c . 4. √ n + a n + b √ n + c . 5. ( √ n + a)(√ n + b) cn + d . 6. cn + d ( √ n + a)(√ n + b) . 7. an3 + b (c √ n + d) 6 . 8. (a √3 n + b) 2 (c √ n + d) 3 . 9. (a √ n + b) 2 (c √3 n + d) 3 . Exercise 1.1.6. Show that limn→∞ apn p + ap−1n p−1 + · · · + a1n + a0 bqnq + bq−1nq−1 + · · · + b1n + b0 = 0, if p < q, ap bq , if p = q and bq 6= 0. Exercise 1.1.7. Find the limits. 1. 1010n n2 − 10 . 2. 5 5 (2n + 1)2 − 1010 10n2 − 5 . 3. 5 5 (2√ n + 1)2 − 1010 10n − 5 . Exercise 1.1.8. Find the limits
1. 1. LIMIT OF SEQUENCE n+a n+c n+b n+d b n2+d n +b td 3+b +b-+d9.+b-边+d Exercise 1.1.9. Find the limits n2+ain+ao n2+cin+ 1 n+b n+d n+ n-+aln+a0 n-+Cin+ co m2+ bin+b0 n2+din+do Exercise 1.1.10. Find the limits, p, q >0 +c +d anP+bn9+c 又.n2p+a1m+a2 anq+bnp+c m2q+61n9+b 1.1.2 Sandwich rule The following property reflects the intuition that if a and z are close to 3, then anything between s and z should also be close to 3 Proposition 1.1.4 (Sandwich Rule). Suppose In Un En for sufficiently big n. If mn→xn=limn→oxn=l, then lim→oyn=l lote that something holds for sufficiently big n is the same as something fails for only finitely many n havinple 1.1.3. By 2n-3 >n for sufficiently big n(in fact, n>3 is enough),we E 0<√2n-3 Then by lin→a0=limn→ 0 and the sandwich rule, we get lim √2n-3 On the other hand, for sufficiently big n, we have n+1 2n and n-1 and therefore vn+I√2n2V2
1.1. LIMIT OF SEQUENCE 11 1. n n + 1 − n n − 1 . 2. n 2 n + 1 − n 2 n − 1 . 3. n √ n + 1 − n √ n − 1 . 4. n + a n + b − n + c n + d . 5. n 2 + a n + b − n 2 + c n + d . 6. n + a √ n + b − n + c √ n + d . 7. n 3 + a n2 + b − n 3 + c n2 + d . 8. n 2 + a n3 + b − n 2 + c n3 + d . 9. √ n + a √3 n + b − √ n + c √3 n + d . Exercise 1.1.9. Find the limits. 1. n 2 + a1n + a0 n + b − n 2 + c1n + c0 n + d . 2. n 2 + a1n + a0 n2 + b1n + b0 − n 2 + c1n + c0 n2 + d1n + d0 . 3. n + a n + b 2 − n + c n + d 2 . 4. n 2 + a n + b 2 − n 2 + c n + d 2 . Exercise 1.1.10. Find the limits, p, q > 0. 1. n p + a nq + b . 2. anp + bnq + c anq + bnp + c . 3. n p + a nq + b − n p + c nq + d . 4. n 2p + a1n p + a2 n2q + b1nq + b2 . 1.1.2 Sandwich Rule The following property reflects the intuition that if x and z are close to 3, then anything between x and z should also be close to 3. Proposition 1.1.4 (Sandwich Rule). Suppose xn ≤ yn ≤ zn for sufficiently big n. If limn→∞ xn = limn→∞ zn = l, then limn→∞ yn = l. Note that something holds for sufficiently big n is the same as something fails for only finitely many n. Example 1.1.3. By 2n − 3 > n for sufficiently big n (in fact, n > 3 is enough), we have 0 < 1 √ 2n − 3 < 1 √ n . Then by limn→∞ 0 = limn→∞ 1 √ n = 0 and the sandwich rule, we get limn→∞ 1 √ 2n − 3 = 0. On the other hand, for sufficiently big n, we have n + 1 < 2n and n − 1 > n 2 , and therefore 0 < √ n + 1 n − 1 < √ 2n n 2 = 2 √ 2 √ n