CHAPTER 1. LIMIT This gives the rigorous proof of limn-oo(n+a-vn)=0 n+ Example 1.213. In Example 1.1.6, we argued that limn-+Vn-1. To make the argument rigorous, we use the estimation in the earlier example. The estimation suggested that it is sufficient to have 2-< E. Thus we get the following rigorous argument for the limit +2 n+2 n E Exercise 1.2.9. Rigorously prove the limits n+2 √n+ 2. +b n+a +1 n+b +a n+c n+a n+b V n+b 1 an+ n+1-y Exercise 1.2. 10. Rigorously prove the limits, p>0 sinn+b nP+b Example 1.2. 14. The estimation in Example 1.1.7 tells us that I va-1 <-for a>1. This suggests that for any e>0, we may choose N=-.Then This rigorously proves that limn-yoo 1 Example 1.2.15. We try to rigorously prove limn-oo n2a"=0 for al<1 Using the idea of Example 1.1.11, we write lal Then jal 1 impl
32 CHAPTER 1. LIMIT This gives the rigorous proof of limn→∞( √ n + a − √ n) = 0. Example 1.2.13. In Example 1.1.6, we argued that limn→∞ r n + 2 n = 1. To make the argument rigorous, we use the estimation in the earlier example. The estimation suggested that it is sufficient to have 2 1 n < . Thus we get the following rigorous argument for the limit n < 2 =⇒ 0 < r n + 2 n − 1 < n + 2 n − 1 = 2 n < =⇒ r n + 2 n − 1 < . Exercise 1.2.9. Rigorously prove the limits. 1. n + 2 n − 3 . 2. n − 2 n + 3 . 3. n + a n + b . 4. 2n 2 − 3n + 2 3n2 − 4n + 1 . 5. √ n + 2 √ n − 3 . 6. √ n + a n + b . 7. n n + 1 − n n − 1 . 8. n + a n + b − n + c n + d . 9. 1 √ an + b . 10. sin √ n n . 11. cos √ n + a n + b sin n . 12. √ n + a − √ n + b. 13. r n + a n + b . 14. √3 n + 1 − √3 n. Exercise 1.2.10. Rigorously prove the limits, p > 0. 1. a √ np + b . 2. n p + a np + b . 3. a sin n + b np + c . Example 1.2.14. The estimation in Example 1.1.7 tells us that | √n a − 1| < a n for a > 1. This suggests that for any > 0, we may choose N = a . Then n > N =⇒ |√n a − 1| < a n < a N = . This rigorously proves that limn→∞ √n a = 1 in case a ≥ 1. Example 1.2.15. We try to rigorously prove limn→∞ n 2a n = 0 for |a| < 1. Using the idea of Example 1.1.11, we write |a| = 1 1 + b . Then |a| < 1 implies
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 6>0. and for n >3. we have 1+mb+n(n-1 n(n-1)(m-2 3! 3!n3 n(7 30=2b (n-1)(n-2份<mn1=mb3 2 3!22 322 Since nb3 E is the same as n> b2∈, we have 3!2 2 andn≥3= This shows that we may choose N= max3! 22 It is clear from the proof that we generally have lim na"=0, for any p and la<1 le 1.2.16. We rigorously prove l1mmn-oo nI=0 in Example 1.1.13 Choose a natural number M satisfying a M. Then for n>M, we have MM·MM M M MM+l 1 m!n!-1.2…MM+1^M+2n MI Therefore for any E>0, we have MM+ M+11 MM+ n> max MIe nl M! n M! MM+I-E Mle Exercise 1.2.1l. Rigorously prove the limits 7. nPan, a<1 m! 4. 1.2.4 Rigorous Proof of Limit Properties The rigorous definition of limit allows us to rigorously prove some limit properties
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 33 b > 0, and for n ≥ 3, we have |n 2 a n − 0| = n 2 |a| n = n 2 (1 + b) n = n 2 1 + nb + n(n − 1) 2! b 2 + n(n − 1)(n − 2) 3! b 3 + · · · + b n < n 2 n(n − 1)(n − 2) 3! b 3 = 3!n (n − 1)(n − 2)b 3 < 3!n n 2 n 2 b 3 = 3!22 nb3 . Since 3!22 nb3 < is the same as n > 3!22 b 3 , we have n > 3!22 b 3 and n ≥ 3 =⇒ |n 2 a n − 0| < 3!22 nb3 < . This shows that we may choose N = max 3!22 b 3 , 3 . It is clear from the proof that we generally have limn→∞ n p a n = 0, for any p and |a| < 1. Example 1.2.16. We rigorously prove limn→∞ a n n! = 0 in Example 1.1.13. Choose a natural number M satisfying |a| < M. Then for n > M, we have a n n! < Mn n! = M · M · · · M 1 · 2 · · · M · M M + 1 · M M + 2 · · · M n ≤ MM M! · M n = MM+1 M! · 1 n . Therefore for any > 0, we have n > max MM+1 M! , M =⇒ a n n! − 0 < MM+1 M! · 1 n < MM+1 M! · 1 MM+1 M! = . Exercise 1.2.11. Rigorously prove the limits. 1. √n n. 2. n 5.4 n! . 3. n p n! . 4. n 5.43 n n! . 5. n pa n n! . 6. n! nn . 7. n pa n , |a| < 1. 1.2.4 Rigorous Proof of Limit Properties The rigorous definition of limit allows us to rigorously prove some limit properties
CHAPTER 1. LIMIT E× ample1.2,17. Suppose lim→sxn=l>0. We prove that lim+san=√ First we clarify the problem. The limit limn-oo In=I means the implication For any E>0, there is N, such that n>N=In-l<e The limit limn→syan=√ l means the implication For isN, such that n>N=→|v We need to argue is that the first implication implies the second implication We have Therefore for any given e>0, the second implication will hold as long as I) or lin-4< Vle. The inequality lan-4< vle can be achieved from the first implication, provided we apply the first implication to vle in place of E. The analysis above leads to the following formal proof. Let E>0. By applying the definition of limn-o n =l to vle >0, there is N, such that N=In-l< vi < =k- In the argument, we take advantage of the fact that the definition of limit can e applied to any positive number, ve for example, instead of the given positive number e Example 1.2. 18. We prove the arithmetic rule limn-oo (an yn )= limn-ooIn+ limn-oo yn in Proposition 1. 1.3. The concrete Example 1.2.9 provides idea of the roof Let limn→∞xn= l and lim→smh=k. Then for any e>0,e2>0,tl N1, N2, such that n> k We expect to choose 61, E2 as some modification of E, as demonstrated in Example .2.17
34 CHAPTER 1. LIMIT Example 1.2.17. Suppose limn→∞ xn = l > 0. We prove that limn→∞ √ xn = √ l. First we clarify the problem. The limit limn→∞ xn = l means the implication For any > 0, there is N, such that n > N =⇒ |xn − l| < . The limit limn→∞ √ xn = √ l means the implication For any > 0, there is N, such that n > N =⇒ |√ xn − √ l| < . We need to argue is that the first implication implies the second implication. We have | √ xn − √ l| = |( √ xn − √ l)(√ xn + √ l)| √ xn + √ l = |xn − l| √ xn + √ l ≤ |xn − l| √ l . Therefore for any given > 0, the second implication will hold as long as |xn − l| √ l < , or |xn − l| < √ l. The inequality |xn − l| < √ l can be achieved from the first implication, provided we apply the first implication to √ l in place of . The analysis above leads to the following formal proof. Let > 0. By applying the definition of limn→∞ xn = l to √ l > 0, there is N, such that n > N =⇒ |xn − l| < √ l. Then n > N =⇒ |xn − l| < √ l =⇒ |√ xn − √ l| = |( √ xn − √ l)(√ xn + √ l)| √ xn + √ l = |xn − l| √ xn + √ l ≤ |xn − l| √ l < . In the argument, we take advantage of the fact that the definition of limit can be applied to any positive number, √ l for example, instead of the given positive number . Example 1.2.18. We prove the arithmetic rule limn→∞(xn + yn) = limn→∞ xn + limn→∞ yn in Proposition 1.1.3. The concrete Example 1.2.9 provides idea of the proof. Let limn→∞ xn = l and limn→∞ yn = k. Then for any 1 > 0, 2 > 0, there are N1, N2, such that n > N1 =⇒ |xn − l| < 1, n > N2 =⇒ |yn − k| < 2. We expect to choose 1, 2 as some modification of , as demonstrated in Example 1.2.17
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT Let N= max(N1, N2 .Then m>N= n>N1, n>N2 ln-lI k (xn+yn)-(+k)≤|xn-l+|yn-k<∈1+2 If E1+E2 < E, then this rigorously proves lim+oo(n+yn)=l+k. Of course this means that we may choose e1=E2=2 at the beginning of the argument The analysis above leads to the following formal proof. For any E>0, apply the definition of limn→∞xn= l and limn→swh=kto>0. We find Ni and N2,sudh n>M1=|n-4<2 n>N2=|9n-k< n>maxN,N→|n-4<2,bm-利<2 →(xn+m)-(+A)≤pn-4+mn-<2+2= Example 1.2. 19. The arithmetic rule limn-+oo Inn= limn-yoo In limn-+oo yn in Propo- sition 1.1.3 means that, if we know the approximate values of the width and height of a rectangle, then multiplying the width and height approximates the area of the rectangle. The rigorous proof requires us to estimate how the approximation of the area is affected by the approximations of the width and height. Example 1.2.11 gives the key idea for such estimation area=Zlly-kI Figure 1. 2.2: The error in product
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 35 Let N = max{N1, N2}. Then n > N =⇒ n > N1, n > N2 =⇒ |xn − l| < 1, |yn − k| < 2 =⇒ |(xn + yn) − (l + k)| ≤ |xn − l| + |yn − k| < 1 + 2. If 1 + 2 ≤ , then this rigorously proves limn→∞(xn + yn) = l + k. Of course this means that we may choose 1 = 2 = 2 at the beginning of the argument. The analysis above leads to the following formal proof. For any > 0, apply the definition of limn→∞ xn = l and limn→∞ yn = k to 2 > 0. We find N1 and N2, such that n > N1 =⇒ |xn − l| < 2 , n > N2 =⇒ |yn − k| < 2 . Then n > max{N1, N2} =⇒ |xn − l| < 2 , |yn − k| < 2 =⇒ |(xn + yn) − (l + k)| ≤ |xn − l| + |yn − k| < 2 + 2 = . Example 1.2.19. The arithmetic rule limn→∞ xnyn = limn→∞ xn limn→∞ yn in Proposition 1.1.3 means that, if we know the approximate values of the width and height of a rectangle, then multiplying the width and height approximates the area of the rectangle. The rigorous proof requires us to estimate how the approximation of the area is affected by the approximations of the width and height. Example 1.2.11 gives the key idea for such estimation. area= |l||y − k| area= |x − l||y| l k x y Figure 1.2.2: The error in product
CHAPTER 1. LIMIT Let limn-oo In=I and limn-oo yn= k. Then for any 61>0, E2>0, there are 1,/2, such that n>N1= kl< es Then for n>N= max(N1, N2), we have(see Figure 1.2.2) J -lk=(an-D)yn+l(gn-k) <laIn-llIyn +l'lIyn-kl ∈1(k|+e2)+|l where we use Iyn -k E2 implying yn <k +E2. The proof of limn-oo TnUn= Lk will be complete if, for any e>0, we can choose 61>0 and E2>0, such that ∈1(k|+62)+|e2≤e This can be achieved by choosing E1, E2 satisfying e2≤1,a(k+1)≤5,|e2≤ n other words, if we choose 2(k+1) ∈=mn 2 t the very beginning of the proof, then we get a rigorous proof of the arithmetic rule. The formal writing of the proof is left to the reader Example 1.2.20. The sandwich rule in Proposition 1.1.4 reflects the intuition that, if and z are within e of 5, then any number y between and z is also within e of 5 <e T Geometrically, this means that if x and z lies inside an interval, say(5-6, 5+e) then any number y between and z also lies in the interval Suppose a≤whn≤ zn and lim→xn=limn→∞an=l. For any e>0, there are Ni and N2, such that lEn-ll Then n>N=max{1,N2}→|xn-l<e,|zn-l<∈ →l-∈<xn,zn<l+∈
36 CHAPTER 1. LIMIT Let limn→∞ xn = l and limn→∞ yn = k. Then for any 1 > 0, 2 > 0, there are N1, N2, such that n > N1 =⇒ |xn − l| < 1, n > N2 =⇒ |yn − k| < 2. Then for n > N = max{N1, N2}, we have (see Figure 1.2.2) |xnyn − lk| = |(xn − l)yn + l(yn − k)| ≤ |xn − l||yn| + |l||yn − k| < 1(|k| + 2) + |l|2, where we use |yn − k| < 2 implying |yn| < |k| + 2. The proof of limn→∞ xnyn = lk will be complete if, for any > 0, we can choose 1 > 0 and 2 > 0, such that 1(|k| + 2) + |l|2 ≤ . This can be achieved by choosing 1, 2 satisfying 2 ≤ 1, 1(|k| + 1) ≤ 2 , |l|2 ≤ 2 . In other words, if we choose 1 = 2(|k| + 1), 2 = min 1, 2|l| at the very beginning of the proof, then we get a rigorous proof of the arithmetic rule. The formal writing of the proof is left to the reader. Example 1.2.20. The sandwich rule in Proposition 1.1.4 reflects the intuition that, if x and z are within of 5, then any number y between x and z is also within of 5 |x − 5| < , |z − 5| < , x ≤ y ≤ z =⇒ |y − 5| < . Geometrically, this means that if x and z lies inside an interval, say (5 − , 5 + ), then any number y between x and z also lies in the interval. Suppose xn ≤ yn ≤ zn and limn→∞ xn = limn→∞ zn = l. For any > 0, there are N1 and N2, such that n > N1 =⇒ |xn − l| < , n > N2 =⇒ |zn − l| < . Then n > N = max{N1, N2} =⇒ |xn − l| < , |zn − l| < =⇒ l − < xn, zn < l + =⇒ l − < xn ≤ yn ≤ zn < l + ⇐⇒ |yn − l| <