1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 1 For any e>0, choose N=-. Then n >N= We need to be more specific on the logical foundation for the arguments. We will assume the basic knowledge of real numbers which are the four arithmetic operations +y, r-g, ry, - the exponential operation y(for >0), the order < y(or y>T, and a s y means a y or a= y), and the properties for these operations. For example, we assume that we already know z>>0 mpliesx<y and cP> yP for p>0. These properties are used in the example above More important about the knowledge assumed above is the knowledge that are not assumed and therefore cannot be used until after the knowledge is established In particular, we do not assume any knowledge about the logarithm. The logarithm and its properties will be rigorously established in Example 1.7. 15 as the inverse of Example 1.2.4. To rigorously prove limn+00 n2+I 1, for any E>0, we have 1 2 n2+1 n2+1N2+1 1+1 Therefore the sequence converges to 1 How did we choose N=,/2 -1? We want to achieve 1 <E. Since this is equivalent to <E, which we can solve to get n 1. choosing 1 should work Example 1.2.5. To rigorously prove the limit in Example 1.1.5, we estimate the difference between the sequence and the expected limit n-0 (n+2)-n /n+2 This shows that for any E >0, it is sufficient to have <∈,Orn> n 22. In othe words, we should choose n
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 27 For any > 0, choose N = 1 1 p . Then n > N =⇒ 1 np − 0 = 1 np < 1 Np = . We need to be more specific on the logical foundation for the arguments. We will assume the basic knowledge of real numbers, which are the four arithmetic operations x + y, x − y, xy, x y , the exponential operation x y (for x > 0), the order x < y (or y > x, and x ≤ y means x < y or x = y), and the properties for these operations. For example, we assume that we already know x > y > 0 implies 1 x < 1 y and x p > yp for p > 0. These properties are used in the example above. More important about the knowledge assumed above is the knowledge that are not assumed and therefore cannot be used until after the knowledge is established. In particular, we do not assume any knowledge about the logarithm. The logarithm and its properties will be rigorously established in Example 1.7.15 as the inverse of exponential. Example 1.2.4. To rigorously prove limn→∞ n 2 − 1 n2 + 1 = 1, for any > 0, we have n > N = r 2 − 1 =⇒ n 2 − 1 n2 + 1 − 1 = 2 n2 + 1 < 2 N2 + 1 = 2 2 − 1 + 1 = . Therefore the sequence converges to 1. How did we choose N = r 2 − 1? We want to achieve n 2 − 1 n2 + 1 − 1 < . Since this is equivalent to 2 n2 + 1 < , which we can solve to get n > r 2 − 1, choosing N = r 2 − 1 should work. Example 1.2.5. To rigorously prove the limit in Example 1.1.5, we estimate the difference between the sequence and the expected limit | √ n + 2 − √ n − 0| = (n + 2) − n √ n + 2 + √ n < 2 √ n . This shows that for any > 0, it is sufficient to have 2 √ n < , or n > 4 2 . In other words, we should choose N = 4 2
CHAPTER 1. LIMIT The discussion above is the analysis of the problem, which you may write on your scratch paper. The formal rigorous argument you are supposed to present is the following: For any E>0, choose N=5 Then n> M +2+√=6 1.2.2 The art of estimation In Examples 1. 2.A, the formula for N is obtained by solving n2+I1<ein exact way. However, this may not be so easy in general. For example, for the limit in Example 1. 1.2, we need to solve 3n-2 While the exact solution can be found, the formula for N is rather complicated. For more complicated example, it may not even be possible to find the formula for the exact solution We note that finding the exact solution of lIn -l< e is the same as finding N=Ne, such that n> However, in order to rigorously prove the limit, only direction is needed. The weaker goal can often be achieved in much simpler way Example 1.2.6. Consider the limit in Example 1. 2. For n> 1, we have 2m2+n 3m-2 3 m+1 n2-n+1n2 2n2+n <∈mmp <E,and、3 n2-n+1 E is equivalen n>-+l, we find that choosing n 3 +1 is sufficient 2n2+n 3n-2 3 3 n>N=-+1→ +1 Exercise 1.2.1. Show that -1<= and then rigorously prove limn→∞n2+ The key for the rigorous proof of limits is to find a simple and good enough stimation. We emphasize that there is no need to find the best estimation. Any estimation that can fulfill the rigorous definition of limit is good enough
28 CHAPTER 1. LIMIT The discussion above is the analysis of the problem, which you may write on your scratch paper. The formal rigorous argument you are supposed to present is the following: For any > 0, choose N = 4 2 . Then n > N =⇒ |√ n + 2 − √ n − 0| = 2 √ n + 2 + √ n < 2 √ n < 2 √ N = . 1.2.2 The Art of Estimation In Examples 1.2.4, the formula for N is obtained by solving n 2 − 1 n2 + 1 − 1 < in exact way. However, this may not be so easy in general. For example, for the limit in Example 1.1.2, we need to solve 2n 2 + n n2 − n + 1 − 2 = 3n − 2 n2 − n + 1 < . While the exact solution can be found, the formula for N is rather complicated. For more complicated example, it may not even be possible to find the formula for the exact solution. We note that finding the exact solution of |xn − l| < is the same as finding N = N(), such that n > N ⇐⇒ |xn − l| < . However, in order to rigorously prove the limit, only =⇒ direction is needed. The weaker goal can often be achieved in much simpler way. Example 1.2.6. Consider the limit in Example 1.1.2. For n > 1, we have 2n 2 + n n2 − n + 1 − 2 = 3n − 2 n2 − n + 1 < 3n n2 − n = 3 n − 1 . Since 3 n − 1 < implies 2n 2 + n n2 − n + 1 − 2 < , and 3 n − 1 < is equivalent to n > 3 + 1, we find that choosing N = 3 + 1 is sufficient n > N = 3 + 1 =⇒ 2n 2 + n n2 − n + 1 − 2 = 3n − 2 n2 − n + 1 < 3 n − 1 < 3 N − 1 = . Exercise 1.2.1. Show that n 2 − 1 n2 + 1 − 1 < 2 n and then rigorously prove limn→∞ n 2 − 1 n2 + 1 = 1. The key for the rigorous proof of limits is to find a simple and good enough estimation. We emphasize that there is no need to find the best estimation. Any estimation that can fulfill the rigorous definition of limit is good enough
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT Everyday life is full of good enough estimations. Mastering the art of such estimations is very useful for not just learning calculus, but also for making smart judgement in real life Example 1.2.7. If a bottle is 20% bigger in size than another bottle, how much bigger Is In The exact formula is the cube of the comparison in size (1+0.2)3=1+30.2+3.0.22+0.23 Since 3.0.2=0.6.3.0.22=0.12 and 0.23 is much smaller than 0.1. the bottle is a little more than 72% bigger in volume Example 1.2.8. The 2013 GDP per capita is 9, 800USD for China and 53, 100USD for the United States , In terms of PPP (purchasing power parity ) The percentage of the annual gdP growth for the three years up to 2013 are 9.3, 7.7, 7.7 for China and 1.8, 2.8, 1.9 for the United States. What do we expect the number of years for China to catch up to the United States? First we need to estimate how much faster is the Chinese gdp growing compared to the United States. The comparison for 2013 is 1+0.077 1+0.019≈1+0.077-0.019)=1+0058 Similarly, we get the(approximate) comparisons 1+0.075 and 1+0.049 for the other two years. Among the three comparisons, we may choose a more conservative 1+0.05. This means that we assume Chinese gdp per capita grows 5% faster than the United States for the next many years Based on the assumption of 5%, the number of years n for China to catch up to the United States is obtained exactly by solving (1+0.05)=1+m0.05+ 0.052+…+0.0 9,800≈5.5 If we use 1+n0.05 to approximate(1+0.05)", then we get n x 2.o However, 90 years is too pessimistic because for n=90, the third ter(.05= 90 0.05 is quite sizable, so that 1+n0.05 is not a good approximation of (1+0.05) getting better estimation. By 53, 100 mation, we try to avoid using calculator in Anan exercise for the art of est ≈232, we may solve n=2m,(1+0.05)m=1+m0.05+ m(m 0.052+…+0.05m≈23
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 29 Everyday life is full of good enough estimations. Mastering the art of such estimations is very useful for not just learning calculus, but also for making smart judgement in real life. Example 1.2.7. If a bottle is 20% bigger in size than another bottle, how much bigger is in volume? The exact formula is the cube of the comparison in size (1 + 0.2)3 = 1 + 3 · 0.2 + 3 · 0.2 2 + 0.2 3 . Since 3 · 0.2 = 0.6, 3 · 0.2 2 = 0.12, and 0.2 3 is much smaller than 0.1, the bottle is a little more than 72% bigger in volume. Example 1.2.8. The 2013 GDP per capita is 9,800USD for China and 53,100USD for the United States, in terms of PPP (purchasing power parity). The percentage of the annual GDP growth for the three years up to 2013 are 9.3, 7.7, 7.7 for China and 1.8, 2.8, 1.9 for the United States. What do we expect the number of years for China to catch up to the United States? First we need to estimate how much faster is the Chinese GDP growing compared to the United States. The comparison for 2013 is 1 + 0.077 1 + 0.019 ≈ 1 + (0.077 − 0.019) = 1 + 0.058. Similarly, we get the (approximate) comparisons 1 + 0.075 and 1 + 0.049 for the other two years. Among the three comparisons, we may choose a more conservative 1 + 0.05. This means that we assume Chinese GDP per capita grows 5% faster than the United States for the next many years. Based on the assumption of 5%, the number of years n for China to catch up to the United States is obtained exactly by solving (1 + 0.05)n = 1 + n 0.05 + n(n − 1) 2 0.052 + · · · + 0.05n = 53, 100 9, 800 ≈ 5.5. If we use 1 + n 0.05 to approximate (1 + 0.05)n , then we get n ≈ 5.5 − 1 0.05 = 90. However, 90 years is too pessimistic because for n = 90, the third term n(n − 1) 2 0.052 is quite sizable, so that 1 + n 0.05 is not a good approximation of (1 + 0.05)n . An an exercise for the art of estimation, we try to avoid using calculator in getting better estimation. By 53, 100 9, 800 ≈ 2.3 2 , we may solve n = 2m, (1 + 0.05)m = 1 + m 0.05 + m(m − 1) 2 0.052 + · · · + 0.05m ≈ 2.3
CHAPTER 1. LIMIT We get n≈2.2.3-1 m(m-1 nce 0.05 2 0.05 is still sizable for m= 26(but giving much better approximation than n= 90), the actual n should be somewhat smaller than 52. We try n= 40 and estimate(1+0.05)" by the first three terms (1+005≈1+40.0.05140.39032≈5 So it looks like somewhere between 40 and 45 is a good estimation We conclude that, if Chinese GDP per capita growth is 5%(a very optimistic assumption) faster than the United States in the next 50 years, then China will catch up to the United States in 40 some years Exercise 1.2.2. I wish to paint a wall measuring 3 meters tall and 6 meters wide, take 10% in each direction. If the cost of paint is $13.5 per square meters, how should i pay for th Exercise 1.2.3. In a supermarket I bought four items at $5.95, $6.35, $15.50, $7.20. The sales tax is 8%. The final bill is around $38. Is the bill correct? Exercise 1.2.4. In 1900, Argentina and Canada had the same GDP per capita. In 2000, the gDP per capita is 9, 300USD for Argentina and 24, 000USD for Canada. On average, how much faster is Canadian gDP growing annually compared with Argentina in the 20th Next we leave real life estimations and try some examples in calculus Example 1.2.9. If z is close to 3 and y is close to 5, then 2z-3y is close to 2.3-3.5 9. We wish to be more precise about the statement, say, we want to find a tolerance for T and y, such that 2m-3y is within +0.2 of-9 We have (2x-3y)-(-9)=|2(x-3)-3(y-5川≤2|x-3+3y-5 For the difference to be within +0. 2, we only need to make sure 2Jz-3+3ly-5< 0. 2. This can be easily achieved by lr-3<0=0.04 and 1y-5<0.04 Example 1.2.10. Again we assume a and y are close to 3 and 5. Now we want to find the percentage of tolerance, such that 2c-3y is within +0.2 of -9 We vainly use the in Example 1. 2.9 and find the percentage 1.33% for a and 0.04 N 0.8% for y. This implies that, if both r and y are within 0.8% of 3 and 5, then 2c-3y is within +0. 2 of-9 The better(or more honest) way is to directly solve the problem. Let d1 and be the percentage of tolerance for r and y. Then = 3(1+51) and y=5(1+d2) (2x-3y)-(-9)=|2(x-3)-3(y-5)|≤|2.361-3562≤215,6=max{|b1,|62
30 CHAPTER 1. LIMIT We get n ≈ 2 · 2.3 − 1 0.05 = 52. Since m(m − 1) 2 0.052 is still sizable for m = 26 (but giving much better approximation than n = 90), the actual n should be somewhat smaller than 52. We try n = 40 and estimate (1 + 0.05)n by the first three terms (1 + 0.05)40 ≈ 1 + 40 · 0.05 + 40 · 39 2 0.052 ≈ 5. So it looks like somewhere between 40 and 45 is a good estimation. We conclude that, if Chinese GDP per capita growth is 5% (a very optimistic assumption) faster than the United States in the next 50 years, then China will catch up to the United States in 40 some years. Exercise 1.2.2. I wish to paint a wall measuring 3 meters tall and 6 meters wide, give or take 10% in each direction. If the cost of paint is $13.5 per square meters, how much should I pay for the paint? Exercise 1.2.3. In a supermarket, I bought four items at $5.95, $6.35, $15.50, $7.20. The sales tax is 8%. The final bill is around $38. Is the bill correct? Exercise 1.2.4. In 1900, Argentina and Canada had the same GDP per capita. In 2000, the GDP per capita is 9,300USD for Argentina and 24,000USD for Canada. On average, how much faster is Canadian GDP growing annually compared with Argentina in the 20th century? Next we leave real life estimations and try some examples in calculus. Example 1.2.9. If x is close to 3 and y is close to 5, then 2x−3y is close to 2·3−3·5 = −9. We wish to be more precise about the statement, say, we want to find a tolerance for x and y, such that 2x − 3y is within ±0.2 of −9. We have |(2x − 3y) − (−9)| = |2(x − 3) − 3(y − 5)| ≤ 2|x − 3| + 3|y − 5|. For the difference to be within ±0.2, we only need to make sure 2|x−3|+ 3|y −5| < 0.2. This can be easily achieved by |x − 3| < 0.2 2+3 = 0.04 and |y − 5| < 0.04. Example 1.2.10. Again we assume x and y are close to 3 and 5. Now we want to find the percentage of tolerance, such that 2x − 3y is within ±0.2 of −9. We can certainly use the answer in Example 1.2.9 and find the percentage 0.04 3 ≈ 1.33% for x and 0.04 5 ≈ 0.8% for y. This implies that, if both x and y are within 0.8% of 3 and 5, then 2x − 3y is within ±0.2 of −9. The better (or more honest) way is to directly solve the problem. Let δ1 and δ2 be the percentage of tolerance for x and y. Then x = 3(1 + δ1) and y = 5(1 + δ2), and |(2x−3y)−(−9)| = |2(x−3)−3(y−5)| ≤ |2·3δ1−3·5δ2| ≤ 21δ, δ = max{|δ1|, |δ2|}
1. 2. RIGOROUS DEFINITION OF SEQUENCE LIMIT To get 21 8 to be within our target of 0.2, we may take our tolerance 8=0.9%< 是≈0.00 Example 1.2.11. Assume r and y are close to 3 and 5 15.n to find the tolerance for a and y, such that y is within +0. 2 of 3 means finding 8>0 such that x-3<,-5<6→|ry-15l<0.2 Under the assumptions -3 <8 and ly-5| <8, we have y-15≤|xy-3+|3y-15≤|x-3ly+3y-5≤(|+3)6 We also note that, if we postulate 8< 1, then Iy-5 8 implies 4 <y< 6, so that xy-15≤(+3)6≤(6+3)0=96 To get 98 to be within our target of 0.2, we may take our tolerance 6=0.02< Since this indeed satisfies s< 1. we conclude that we can take d=0.02 If the targeted error +0.2 is changed to some other amount te, then the same argument shows that we can take the tolerance to be d= f. Strictly speaking, since we also use 8< l in the argument above, we should take 8=min 1, 11 Exercise 1.2.5. Find a tolerance for y, z near-2, 3, 5, such that 5. -3y+ 4z is within Exercise 1.2.6. Find a tolerance for a and y near 2 and 2, such that y is within te of 4. Exercise 1.2.7. Find a tolerance for near 2 such that 2 is within +e of 4 Exercise 1. 2.8. Find a percentage of tolerance for a near 2, such that -is within +0. 1 of 1.2. 3 Rigorous proof of limits We revisit the limits derived before and make the argument rigorous Example 1.2.12. In Example 1.1.5, we argued that limn-oo(n make the argument rigorous, we use the estimation in the earlier example. In fact, regardless of the sign of a, we always have n-0 mn+a-Vm)(√m+a+vm n+a+vn For the right side <E, it is sufficient to have n >5. Then we can easily get →|Vn+a-√n-0<∈
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 31 To get 21|δ| to be within our target of 0.2, we may take our tolerance δ = 0.9% < 0.2 21 ≈ 0.0095. Example 1.2.11. Assume x and y are close to 3 and 5. We want to find the tolerance for x and y, such that xy is within ±0.2 of 3 · 5 = 15. This means finding δ > 0, such that |x − 3| < δ, |y − 5| < δ =⇒ |xy − 15| < 0.2. Under the assumptions |x − 3| < δ and |y − 5| < δ, we have |xy − 15| ≤ |xy − 3y| + |3y − 15| ≤ |x − 3||y| + 3|y − 5| ≤ (|y| + 3)δ. We also note that, if we postulate δ ≤ 1, then |y − 5| < δ implies 4 < y < 6, so that |xy − 15| ≤ (|y| + 3)δ ≤ (6 + 3)δ = 9δ. To get 9|δ| to be within our target of 0.2, we may take our tolerance δ = 0.02 < 0.2 9 . Since this indeed satisfies δ ≤ 1, we conclude that we can take δ = 0.02. If the targeted error ±0.2 is changed to some other amount ±, then the same argument shows that we can take the tolerance to be δ = 10 . Strictly speaking, since we also use δ ≤ 1 in the argument above, we should take δ = min{ 10 , 1}. Exercise 1.2.5. Find a tolerance for x, y, z near −2, 3, 5, such that 5x − 3y + 4z is within ± of 1. Exercise 1.2.6. Find a tolerance for x and y near 2 and 2, such that xy is within ± of 4. Exercise 1.2.7. Find a tolerance for x near 2, such that x 2 is within ± of 4. Exercise 1.2.8. Find a percentage of tolerance for x near 2, such that 1 x is within ±0.1 of 0.5. 1.2.3 Rigorous Proof of Limits We revisit the limits derived before and make the argument rigorous. Example 1.2.12. In Example 1.1.5, we argued that limn→∞( √ n + a − √ n) = 0. To make the argument rigorous, we use the estimation in the earlier example. In fact, regardless of the sign of a, we always have | √ n + a − √ n − 0| = ( √ n + a − √ n)(√ n + a + √ n) √ n + a + √ n = |a| √ n + a + √ n < |a| √ n . For the right side |a| √ n < , it is sufficient to have n > a 2 2 . Then we can easily get n > a 2 2 =⇒ |√ n + a − √ n − 0| <