Limits and Continuity

Chapter Limits and Continuity. Limits of Sequences.. The Concept of Limit and Its Properties A sequence { } is an ordered infinite list x,x,...,,... The n-th term of the sequence is, and n is the index of the term. In this course, we will always assume that the terms are real numbers. Here are some examples of sequences of real numbers = n,, 3,..., n,... ; y n =,,,...,,... ; z n = n,,..., n,... ; u n = ( ) n,,,...,( ) n,... ; v n = sin n sin, sin, sin 3,...,sin n,.... Note that the index does not have to start from. For example, the sequence {u n } actually starts at 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 {w n }: w n 3, 3., 3.4, 3.4, 3.45, 3.459, 3.459,.... If n is the number of digits after the decimal point in w n, then the sequence {w n } starts from n = 0. We can examine the behaviors of these sequences as n becomes larger and larger. Apparently, in this case, the term gets larger. In fact, can be larger than any fixed number if the inde is sufficiently large. On the other hand, y n remains constant, and z n can be smaller than any fixed positive number if n is sufficiently large. Our intuition tells that y n approaches and z n approaches 0. Moreover, the terms u n and v n jump around and do not approach any number. Finally, w n gets closer and closer to π. In fact, w n is equal to π up to the n-th decimal point. 3

4 Chapter. Limits and Continuity w n y n u n vn z n n Figure.: Examples of several sequences. Therefore, as n gets larger, the sequences {y n }, {z n }, {w n } approach finite numbers, 0, and π, while the sequences { }, {u n }, {v n } do not approach any specific finite numbers. These sequences are illustrated in Figure.. Definition.. (Non-rigorous) If approaches a finite number l when n gets larger and larger (or goes to infinity), then we say that the sequence { } converges to the it l, or simply that exists, and write = l. A sequence diverges if it does not converge to any specific finite number when n goes to infinity. Since the it describes the behavior when n gets very large, the modification of finitely many terms in a sequence changes neither the convergence nor the it value. Proposition.. If exists, and if {y n } is obtained from { } by adding, deleting, or changing finitely many terms, then y n exists, and y n =. Example.. The sequence { } is obtained from { } by deleting the first n + n two terms. By n = 0 and Proposition.., we get n = = 0. n + In general, the same reason tells us that if exists, then +k = for any integer k. Intuitively, we know that 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 = 5. These lead to the following properties.

.. Limits of Sequences 5 Proposition..3 (Arithmetic Rule) Suppose = l and y n = k. Then the arithmetic combinations of the sequences converge, and ( + y n ) = l + k, (c ) = c l, ( y n ) = k l, = l y n k, where c is a constant and k 0 in the last equality. The proposition tells us that ( + y n ) = + y n under the condition that both and y n exist. However, if this condition fails, then in general we cannot conclude that ( + y n ) = + y n. For instance, if = ( ) n and y n = ( ) n+, then ( + y n ) = 0, and both and y n diverge. We therefore have ( + y n ) + y n. Example.. By c = c, = 0, and the arithmetic rule, we have n n + n n n + = = + n n + n = + n n + ( + n ) ( n + n ) n n =. 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..4 (Sandwich Rule) If y n z n and = z n = l, then y n = l. Example..3 We have n ( )n n n, n sin n n n. Moreover, we have n = = 0. By the arithmetic rule, we know n ( ) = n ( ) = 0. Therefore we get n ( ) n sin n = n n = 0.

6 Chapter. Limits and Continuity The argument for the first it can be generalized as follows: If = 0, then = 0. See Exercise..6. Example..4 Since 0 < n + n = ( n + n)( n + + n) = <, n + + n n + + n n by the sandwich rule and we get = n = 0, n ( n + n) = 0. Example..5 We show that n a =, for a > 0. First assume a. Put = n a. Then 0, and a = ( + ) n = + n + n(n ) x n + + n n. This gives us 0 a n. By the sandwich rule and arithmetic rule, a n = a n = 0, we get = 0. n a = ( + ) = + =. Hence, by the Next, in the case 0 < a <, we put b =. Then b >. By the above discussion, and a the arithmetic rule, we have n a = n = b n =. b

.. Limits of Sequences 7 Example..6 The idea of Example..5 can also be used to show that n n =. Let = n n. Then 0, and This implies n = ( + ) n = + n + n(n ) x n + + n 0 n. n(n ) x n. By the sandwich rule and Example.., we get = 0. It follows that n n = ( + ) =. Example..7 The following sequences of n-th root type can be compared with the sequences in Examples..5 and..6. < n n + < n n = n n n, < n n+ < n n, < (n n) n n < (n ) n n < (n ) n n / = ( n n) 4. By Examples..6,..7, and the arithmetic rule, the sequences on the right hand side all converge to. Then by the sandwich rule, we have n n + = n n+ = (n n) n n =. Example..8 We have 3 = n 3 n < n n + 3 n < n 3 n = 3 n. By Example..5, the it of the right hand side is 3. By the sandwich rule, we get n n + 3 n = 3. Example..9 We show an = 0, for a <.

8 Chapter. Limits and Continuity First assume 0 < a < and write a =. Then b > 0 and + b 0 < a n = ( + b) n = n(n ) + nb + b + + b n nb. By = 0, and the sandwich rule, we get nb an = 0. Next, if < a < 0, then 0 < a < and a n = a n = 0. By the remark in Example..3, we get a n = 0. Example..0 We have a n = 0, for any a. n! Here we discuss the special case with a = 5 (the general case being similar). For n > 5, we have 0 < 5n n! = 5 5 5 5 5 3 4 5 5 6 5 7 5 n 5 5 5 5 5 3 4 5 5 n = 56 5! n. 5 6 By 5! n = 56 5 n = 0, and the sandwich rule, we get 5! n n! = 0. The following property reflects the intuition that larger sequence should have larger it. Proposition..5 (Order Rule) Suppose and y n exist.. If y n for sufficiently large n, then y n.. If > y n, then > y n for sufficiently large n. Note that a property fails for only finitely many n means that it holds for sufficiently large n. By taking y n = k, we get the following properties for a convergent sequence { }.. If k for sufficiently large n, then k.. If > k, then > k for sufficiently large n. Similar statements can be made by taking to be constant. Example.. We must pay special attention to the difference between inequality and strict inequality in the two statements of Proposition..5. For example, although = n > y n =, we have n = y n.

.. Limits of Sequences 9 n + n Example.. By n =, we know < n + n n + n < 3 for sufficiently n + large n, by Statement in Proposition..5. This implies < n n + n n n + < n 3. By n n n + n 3 = and the sandwich rule, we get n n + =. In general, the argument tells us that, if = l > 0, then n =. Example..3 The sequence = 3n (n!) (n)! satisfies 3n = n(n ) = 3 4 = 0.75. Therefore by the order rule, we have < 0.8 for sufficiently large n, say for n > N for a fixed big natural number N (in fact, N = 8 is enough). Then for n > N, we have 0 < = x N+ x N x N < 0.8 n N x N = C 0.8 n, C = 0.8 N x N. By Example..9, we have 0.8 n = 0. Since C is a constant, by the sandwich rule, we get = 0. In general, the argument shows that, if = l and l <, then = 0. Exercises.. Find the its of the sequences (assuming we already know n = n = 0):. n ; 5. 3 n + an+ b ;. n + n ; 6. sin n! n ; 3. n 3 ; 7. ( ) n (n + ) n + ( ) n+ ; 4. 3 n n + ; 8. 5 5 (n + ) 0 0 ; 0n 5

0 Chapter. Limits and Continuity 9. 0... 3. 0 0 n n 0 ; n n! ; n + 3n + 5 ; n! n + n 3 n + 5! ; n! n + n! + (n )! 3 n n! + (n )! ; 4. n + sin n n cos n; 5. n + a n + b; 6. n + + n + n + 3; 7. 3 n + 3 n + ; n 8. n + ; 3 n 9. n + ; n + a 0. n + b ; n + n +. n n + ;. a0 n + a n + a, b 0 n + b n + b a 0,b 0 > 0; 3. a0 n + a n + a, b 0 n + b a 0 > 0,b 0 0; 4. n( n + n); 5. n + a( n + b n + c); 6. n + n + 3 n n + ; 7. n + a n + a n + b n + b ; 8. n( n + n + 3 n + n ); 3 9. n ( 3 n + 3 n); 30. ( n + a π n + b ) ; 3. ( n π + a n + a ) ; n + b n + b 3. ( a 0 n p + a n + a ), a b 0 n 0,b 0 > 0; + b n + b 33. (n + ) n ; 34. (n + sin n) 35. n 7 n ; 36. n 7 + n ; 37. n 7n + n ; n 38. 7 n+ + n ; n 39. 7n + n+ ; n 40. 7n n+ ; 4. n n n ; 4. n n + n ; n 43. (n + ) n ; 44. n n+cos n ; n 3n + 4 n+ + 3n n... First show that Then find the its of the following sequences:.. n + n ; n n ; 3. n 99 0.99 n ; nan = 0, for a <. 4. n + n 3 ; n 5. n 3 n ( + n ) ; 6. 5 n n 6 n+ 3 n 3n+ ;..3 Use = 0 for p > 0 to find the following its: np. 3 n + 3 n; n + 3. 3 n n + ; 7. n 3 a n, a < ; 8. n p a n, a < ; 9. n p, a >. an. 3 n + 3 n ; 4. n + sin n n cos n.

.. Limits of Sequences..4 Explain that = L if and only if ( L) = 0. Use this observation to derive n + = from Example..4. n..5 If ( y n ) = 0, can you conclude that either = 0 or y n = 0?..6 Use the sandwich rule to prove that = 0 implies = 0. Is the converse true?..7 Prove that if converges to a positive number, then n =...8 Prove that if c for a constant c <, then = 0...9 Prove that if = l and l <, then = 0. Use this and Example..9 to derive the it in Example..0 for any a. Moreover, what happens when l?..0 Prove that if n = l and l <, then = 0. What happens if l?.. Bounded Sequences A sequence { } is bounded above if B for a constant B (called an upper bound) and all n. Similarly, it is bounded below if B for a constant B (called a lower bound) and all n. If a sequence is bounded above and below, then we say it is bounded. The order rule implies the following. Proposition..6 Any convergent sequence is bounded. For example, the sequences {n} and { n + ( ) n } diverge because they are not n + bounded above. On the other hand, a bounded sequence does not have to converge, as shown by the sequence,,,,... Therefore the converse of Proposition..6 does not hold in general. The converse of Proposition..6 does hold under some additional assumption. A sequence { } is increasing if It is strictly increasing if x x x 3 +. x < x < x 3 < < < + <. The concepts of decreasing and strictly decreasing can be similarly defined. Moreover, a sequence is monotonic if it is either increasing or decreasing. The sequences { n }, { n }, { n } are (strictly) decreasing. The sequences { n }, {n} are increasing. Proposition..7 Any bounded monotonic sequence converges. An increasing sequence { } is bounded if it is bounded above, because the first term of the sequence is already the lower bound. Similar remark can be made for a decreasing sequence.

Chapter. Limits and Continuity Example..4 The sequence, +, + +,... is recursively given by x =, + = +. We claim that the sequence { } is increasing and bounded above by. We have x = < and x = + > x. Assume < and + >. Then + = + < + =, + = + + > + = +. The claim is therefore proved by induction. By Proposition..7, the sequence converges. Let l be the it. Then by taking the its on both sides of the equality x n+ = + and applying the arithmetic rule, we get l = + l. The solution is l = or. Since > 0, by the order rule we must have l 0. Therefore we conclude that =. Figure. schematically demonstrates the iting process of a recursively defined sequence. y y = x f(x 3 ) y = f(x) f(x ) f(x ) x x x 3 x 4 l x Figure.: Limit of a recursively defined sequence + = f( ). Example..5 We give another argument that a n = 0 for 0 < a <. See Example..9. Since 0 < a <, the sequence {a n } is decreasing and satisfies 0 < a n <. Hence, by Proposition..7, the sequence converges to a it l. By the remark in Example.., we also have a n = l. Then by the arithmetic rule, Since a <, we get l = 0. l = a n = a a n = a a n = al.

.. Limits of Sequences 3 Example..6 We give another argument that the it in Example..3 con- = 0.75 < and the order rule, we have < for sufficiently verges to 0. By large n. Since all terms are positive, we have < for sufficiently large n. Therefore after finitely many terms, the sequence is decreasing. Moreover, 0 is a lower bound of the sequence, so that the sequence converges. Let = l. Then we also have = l. If l 0, then = x = l n l =. But the it is actually 0.75. The contradiction shows that l = 0. Example..7 We show that the sequence {( + n ) n} converges as n. By comparing the binomial expansions of two consecutive terms ( + n n ) = + n n + n(n )! = +! +! ( n ) + n(n ) 3 + + n n! + n! ( n )( n ) ( n n ), ( + n + ) n+ = +! +! ( n + ) + we see that the sequence is increasing: + n! ( n + )( n + ) ( n n + ) + (n + )! ( n + )( n + ) ( n n + ), ( + n n ) ( + n+ n + ). It remains to show that the sequence is bounded above. By the expansion above, we have ( + n n ) < +! +! + + n! < + + + 3 + + (n )n = + + ( ) + ( 3 ) + + ( n n ) = + + n < 3. n n

4 Chapter. Limits and Continuity Hence, by Proposition..7, the sequence converges. The it is denoted by e ( + n n ) = e. It is clear that < e 3 by the order rule. As a matter of fact, it can be shown that e =.7888... is an irrational number. Example..8 Consider the sequence { }: = + + 3 + + n. Moreover, similar to the computation in Exam- This sequence is clearly increasing. ple..7, we have + + 3 + + (n )n = n <. Hence the sequence is bounded, and converges by Proposition..7. The it of the sequence is the sum of the infinite series n= n = + + 3 + + n +, and is actually equal to π 6 (this last fact will not be proved here). Without the condition of monotonicity, a bounded sequence may not converge. The counterexamples, such as,,,,..., suggest that the divergence may be due to the fact that parts of the sequence may still converge, but different parts may converge to different its. Therefore there is no single value for the sequence to converge to. To describe the phenomenon that part of a sequence may converge, we introduce the following concept. A subsequence of a sequence { } is obtained by choosing infinitely many terms and still arranging them in the order as in { }. The subsequence is then,,...,k,..., or {k } in short, with the indices satisfying n < n < < n k <. If the sequence { } starts from n =, then n, which further implies n k k for all k.

.. Limits of Sequences 5 The following are some subsequences of { = n }, corresponding to the choices n k = k, n k = k, n k = k, and n k = k!, respectively: x k = k, 4, 6,, k, ; x k = k, 3, 5,, k, ; x k = k, 4, 8,, k, ; x k! = k!,, 6,, k!,. If something is close to l, then any part of it is also close to l. This leads to the following result. Proposition..8 If a sequence converges to l, then every subsequence converges to l. Example..9 n + n By n =, we also know n + (n ) + (n ) (n ) (n ) + = n 4 n + n 4 3n + 3 =, and (n!) + n! (n!) n! + =. Example..0 Consider the sequence = ( ) n. The subsequence x k = converges to, and the subsequence x k = converges to. Since the two subsequences have different its, Proposition..8 tells us that diverges. Moreover, the expectation that a bounded sequence is made up of a number of convergent subsequences (with perhaps different its) is reflected below. Theorem..9 (Bolzano-Weierstrass) Every bounded sequence has a convergent subsequence. The Bolzano-Weierstrass Theorem is a deep result that touches the essential difference between real numbers and rational numbers.

6 Chapter. Limits and Continuity Example.. Let us list all finite decimal expressions in (0, ) as a sequence { }: 0., 0.,...,0.9, 0.0, 0.0,...,0.99, 0.00, 0.00,...,0.999,... The number 0.38309 is the it of the sequence 0.3, 0.3, 0.38, 0.383, 0.3830, 0.38309,..., which is clearly a subsequence of { }. It is easy to see that any number in [0, ] is the it of a convergent subsequence of { }. Exercises.. Determine the convergence of the following sequences:.. 3. 4. ( ) n n n 3 ; ( ) n n n ; ( ) n n n ; cos n n ; 5. n ( )n ; 6. n n! ; 8. 9. 0... n + cos n n + ( ) n n + sin n ;,, 3, 3, 3 4, 4 3,..., n n +, n + n,...; 3 5 (n ) 4 6 (n) ; 3 5 (n ) ; n! + + 3 + + n ; 7. n! n ; 3. 3 + 3 + 3 3 + + n 3... Determine the convergence of the following recursively defined sequences.. + = +, x > 0;. + =, x > 0; 3. + = a +, a > 0, x > 0; 4. + = ( + a ), a > 0...3 Explain the continued fraction expansion = +. + + + What if on the right hand side is changed to some other positive number?..4 For any a, b > 0, define a sequence by Prove that the sequence converges. x = a, x = b, = +.

.. Limits of Sequences 7..5 The arithmetic and the geometric means of two positive numbers a, b are a + b and ab. By repeating the process, we get two sequences defined by x = a + b, y = ab, + = + y n, y n+ = y n. Prove that + y n+ y n, and the two sequences converge to the same it...6 The Fibonacci sequence,,, 3, 5, 8, 3,, 34,... is defined by x 0 = x = and + = +. Consider the sequence y n = +.. Find the relation between y n+ and y n.. Assume the sequence {y n } converges, find the it l. 3. Use the relation between y n+ and y n to prove that l is an upper bound of y k and a lower bound of y k+. 4. Prove that the subsequence {y k } is increasing and the subsequence {y k+ } is decreasing. 5. Prove that the sequence {y n } converges to l...7 Let = ( + n ) n+.. Use induction to prove ( + x) n + nx for x > and any natural number n.. Use the first part to prove 3. Prove that = e. <. This shows that { } is decreasing. 4. Prove that {( n ) n} is increasing and converges to e...8 Prove that for n > k, we have ( + n n ) +! +! ( n ) + + k! ( n )( n ) ( k n ). Then use Proposition..5 to show that e +! +! + + k! ( + k ) k. Finally, prove ( +! +! + + n! ) = e...3 Rigorous Definition The meaning of the phrases approaching l and going to infinity in the intuitive definition of it is rather ambiguous. Before we introduce the rigorous definition, let

8 Chapter. Limits and Continuity us study the it n = 0 in more detail. When we say that = n approaches l = 0 as n goes to infinity, we mean an infinite collection of facts n > l <, n > 0 l < 0., n > 00 l < 0.0, n > 000000 l < 0.00000, Here, the symbol stands for implies. In everyday life, whether certain quantity is considered large or small depends on the context. For example, the population of millions of people (n > 00000000) is considered as large for a city but rather small as a country. On the other hand, the diameter within one mileter (d < ) is considered thin for an electric wire. But the hair is considered thin only if the diameter is less than 0.05 mileter (d < 0.05). Thus large or small is meaningful only when compared with some reference quantity. For the it n = 0, we say that, if n is in the hundreds, then l is within the hundredth, and if n is in the millions, then l is within the millionth, etc. Of course for a different sequence, the exact relation between the largeness of n and the smallness of l may be different. For example, for = n and l = 0, we have n! n > 0 l < 0.0003, n > 0 l < 0.0000000000005, But the key observation here is that a it is an infinite collection of statements of the form if n is larger than certain large number N, then l is smaller than certain small number ɛ. In practice, we cannot verify all such statements one by one. Even if we have verified the truth of the first one million statements, there is no guarantee that the one million and the first statement is true. To mathematically establish the truth of all such statements, we have to formulate one statement that includes the consideration for all N and ɛ. Which one statement do we need to establish? We note that n > N is the cause of the effect l < ɛ, and the precise logical relation between the cause and the effect varies from it to it. For some it, n in thousands already guarantees that l < 0.00000. For some other it, n has to be in the billions in order to guarantee that l < 0.00000. But the key observation here is that no matter how small ɛ is, the target l < ɛ can always be guaranteed for sufficiently large n. So although l < 0.00000 may not be satisfied for n in the million range, it will probably be satisfied for n in the billions range. In fact, for any given range ɛ > 0, we can always find an integer N such that is close to the it l within the range ɛ whenever n > N. In other words, n > N l < ɛ. For = n, we can take N = ɛ.

.. Limits of Sequences 9 Definition..0 (Rigorous) A sequence { } converges to a finite number l if for any ɛ > 0, there is N, such that n > N implies l < ɛ. In case N is a natural number (which can always be arranged if needed), the condition means that all the terms x N+,x N+,x N+3,... lie in the shaded area in Figure.3. x 4 x x x 3 x N+ x N+xN+3 xn l + ɛ l l ɛ N n Figure.3: n > N implies l < ɛ. Example.. We rigorously prove For any ɛ > 0, choose N =. Then ɛ p = 0, for p > 0. np n > N n p 0 = n p < N p = ɛ. Since we will carry out rigorous arguments, we need to be more specific about their logical foundations. We need to assume the basic knowledge of real numbers. This include. The basic arithmetic operations: x + y, x y, xy, x y.. Exponentiation: for x > 0 and y any real number, we know how to raise x to the y-th power to obtain x y. 3. The order relations: for two real numbers x, y, we know the meaning of x < y and x y, and the properties associated to such inequalities. For example, we assume that we already know x > y > 0 implies x < y and xp > y p for p > 0. These properties have already been used in the argument in the example above. On the other hand, at this point we do not assume that we know anything about the logarithm. The logarithm and its properties will be rigorously established in a later section. Example..3 To rigorously prove the claimed conclusion in Example..4, we estimate the difference between the n-th term of the sequence and the expected it: n + n 0 = (n + ) n n + + n < n.

0 Chapter. Limits and Continuity Therefore for any ɛ > 0, it suffices to have choose N = 4 ɛ. < ɛ, or n > 4. In other words, we should n ɛ The discussion above is an analysis of the problem, which you should probably write on your scratch paper. The following is the formal rigorous argument that you are supposed to present: For any ɛ > 0, choose N = 4 ɛ. Then n > N n + n 0 = < < = ɛ. n + + n n N Example..4 Consider the sequence n n. For any ɛ > 0, we have + n > N = ɛ n n + = n + < N + = ( = ɛ. ɛ ) + Therefore the sequence converges to. How did we choose N =? First note that we want to achieve ɛ This is equivalent to the inequality Therefore, choosing N = n n < ɛ. + n > ɛ. does the trick. ɛ In Examples..4, the formula for N is obtained by solving n n < ɛ in an + exact way. However, this may not be so easy in general. For example, for the it in Example.., we need to solve n + n n n + = 3n n n + < ɛ. 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.

.. Limits of Sequences We note that finding the exact solution of l < ɛ is the same as finding N = N(ɛ), such that n > N l < ɛ. However, in order to rigorously prove that the it is what we claim to be, we only need to establish one direction of the implication, namely. In other words, there is no need to find the exact N. All we need to do is to find one (any) N such that the implication is true. Furthermore, in the statement n > N l < ɛ, the inequality l < ɛ holds under the condition n > N. In other words, l < ɛ is a conditional inequality. We have seen many un-conditional inequalities before. For example, + x and x x + are un-conditional inequalities. Hence, proving = l is to find N so that the inequality l < ɛ holds under the the condition n > N. These observations lead to a so-called loose-and-track approach for proving = l. We demonstrate it in the following examples. Example..5 the inequality To prove n n + =, we need to find a condition under which n n + < ɛ holds. This can be done by loosing up the term on the left hand side in the following sequence of inequalities: n n + = n + < n < N 3 ɛ, where,, and 3 indicates the condition under which the corresponding inequality holds. More explicitly, we have for all natural integers n; n > N; 3 N /ɛ. Thus, the whole sequence of inequalities hold when n > N, with N a number greater than or equal to /ɛ. Hence, by combining,, and 3, we have Therefore, we get n n + =. n > /ɛ n n < ɛ. + Example..6 Example..5 tells us that n a a n n a a n < a N 3 = ɛ, for a. Hence

Chapter. Limits and Continuity where Hence, for a, we have for all natural integers n; n > N; 3 N = a/ɛ. n > a/ɛ n a < ɛ. This rigorously proves that n a = in the case a. Similarly, the estimation in Example..6 can be turned into a rigorous proof of where n n =. In fact, we have n n n < N 3 < N 4 = ɛ, for all natural integers n > ; n > N; 3 for any N > ; 4 N = 4/ɛ +. Hence, by combining all the conditions, we have n > N = 4 ɛ + n n < ɛ. Example..7 We try to rigorously prove n a n = 0 when a <. Using the idea of Example..9, we write a =. Then b > 0, since a <. For + b n 4, we have where n a n 0 = n a n n = ( + b) n = n n(n ) + nb + b n(n )(n ) + b 3 + + b n! 3! n 3! 3 < nb 3 < 3! 4 Nb 3 = ɛ, n(n )(n ) b 3 3! n > ; n 4; 3 n > N; 4 N = 3! b 3 ɛ. < 3!n n n n = b3

.. Limits of Sequences 3 Hence, 4 n > max { ( a ) 3 ɛ, 4} n a n 0 < ɛ, so that n a n = 0 when a <. More generally, it is clear from the proof that we have np a n = 0, for any p and a <. After familiarizing ourselves with the loose-and-track approach, we may skip the intermediate steps to give rigorous proofs, as demonstrated in the following examples. a n Example..8 We rigorously prove = 0 in Example..0 for general a. n! Choose a natural number M, such that a < M. Then for n > M, we have an n! < M n = M M M n! M M M + M M + M n M M M! Therefore for any ɛ > 0, we have n > max { M M+ M+ M,M} an 0 < M!ɛ n! M! n < M M+ M! a n so that = 0 for any fixed a. n! M n = M M+ M! M M+ M!ɛ n. = ɛ, The rigorous definition of it allows us to prove (rigorously) some properties of its. Example..9 Assume = l > 0. We prove that xn = l. The question is the following. We already know We want to show For any ɛ > 0, there is N, such that n > N l < ɛ. For any ɛ > 0, there is N, such that n > N l < ɛ. In short, we need to prove that the first implication implies the second implication. Note that l = ( l)( + l) xn + l = l xn + l l l.

4 Chapter. Limits and Continuity Hence, for any given ɛ > 0, the second implication will hold if l < ɛ, or l < lɛ. l But the inequality l < lɛ follows from the first implication, provided we apply the first implication to lɛ in place of ɛ. The analysis above gives the following formal proof. Let ɛ > 0. By applying the definition of = l to the positive number lɛ > 0, there is N, such that n > N implies l < lɛ. Then n > N l < lɛ l = ( l)( + l) xn + l = l xn + l l l < ɛ. A key point here is to take advantage of the fact that the definition of it can be applied to any positive number, lɛ for example, not necessarily the given positive number ɛ. Example..30 Proposition..3. Let that We prove the arithmetic rule ( + y n ) = + y n in = l and y n = k. Then for any ɛ > 0, ɛ > 0, there are N, N, such n > N l < ɛ, n > N y n k < ɛ. The reason for specifying ɛ,ɛ instead of ɛ is that we expect to choose them as some modification of ɛ, as demonstrated in Example..9. Let N = max{n,n }. Then n > N n > N,n> N l < ɛ, y n k < ɛ ( + y n ) (l + k) l + y n k < ɛ + ɛ. If ɛ + ɛ ɛ, then this rigorously proves ( + y n ) = l + k. Of course this means that we may choose ɛ = ɛ = ɛ at the first step of the argument. The analysis above gives the following formal proof. For any ɛ > 0, apply the definition of = l and y n = k to ɛ > 0. We know that there are N and N, such that Then n > N l < ɛ, n > N y n k < ɛ. n > N = max{n,n } l < ɛ, y n k < ɛ ( + y n ) (l + k) l + y n k < ɛ + ɛ = ɛ.

.. Limits of Sequences 5 Example..3 We prove the arithmetic rule ( y n ) = y n in Proposition..3. Let that = l and y n = k. Then for any ɛ > 0, ɛ > 0, there are N, N, such n > N l < ɛ, n > N y n k < ɛ. Then for n > N = max{n,n }, we have (see Figure.4) y n lk = ( l)y n + l(y n k) l y n + l y n k < ɛ ( k + ɛ ) + l ɛ, where we use y n l < ɛ implies y n < k + ɛ. The proof of ( y n ) = lk will be complete if, for any ɛ > 0, we can choose ɛ > 0 and ɛ > 0, such that ɛ ( k + ɛ ) + l ɛ ɛ. This can be achieved by choosing ɛ,ɛ satisfying ɛ, ɛ ( k + ) ɛ, ( l + )ɛ ɛ. (In the last inequality, we added to l to avoid a possible zero denominator in the following expression of ɛ in the case of l = 0.) Therefore, if we choose ɛ = ɛ ( k + ), ɛ ɛ = min {, ( 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. l k area= xn l yn y n area= l y n k Figure.4: The shaded region represents the error term.

6 Chapter. Limits and Continuity Example..3 We prove the sandwich rule of Proposition..4. Suppose y n z n and = z n = l. For any ɛ > 0, there are N and N, such that n > N l < ɛ, n > N z n l < ɛ. Then n > N = max{n,n } l < ɛ, z n l < ɛ l ɛ <,z n < l + ɛ l ɛ < y n z n < l + ɛ y n l < ɛ. Example..33 We prove Proposition..6, which says that a convergent sequence must be bounded. Let = l. Then for ɛ = > 0, there is N, such that n > N implies l <. By choosing a larger N if necessary, we may further assume that N is a natural number. Then l + is an upper bound of all, n > N. This implies that max{x,...,x N,l+ } is an upper bound of the whole sequence. A lower bound can be found in a similar way. Note that we may always choose N to be a natural number in the definition of it. Such choice is needed here because we also use N as the index of a term in the sequence. Example..34 We prove the order rule in Proposition..5. Assume that y n for sufficiently large n, = l, and y n = k. Let ɛ > 0 be given. Then, there are N,N, and N 3 such that n > N y n ; n > N l < ɛ; n > N 3 y n k < ɛ. Hence, if n > max{n,n,n 3 }, then we have l + ɛ > y n > k ɛ. Since ɛ > 0 is arbitrary, the inequality l + ɛ > k ɛ implies l k; otherwise l < k leads to a contradiction if we choose ɛ = k l. l + k l > k k l,

.. Limits of Sequences 7 Conversely, we assume = l, y n = k, and l > k. For any ɛ > 0, there is N, such that n > N implies l < ɛ and y n k < ɛ. Then n > N > l ɛ, y n < k + ɛ y n > (l ɛ) (k + ɛ) = l k ɛ. By choosing ɛ = l k > 0 at the beginning of the argument, we conclude that > y n for n > N. The choice l k can also be seen from intuition. For large n, is close to l and y n is close to k. If the closeness is within half of the distance between l and k, i.e., within ɛ = l k, then we see that, for large n, is above the midpoint l + k between l and k, and y n is below the midpoint. Example..35 We prove Proposition..8 about the it of subsequences. Suppose = l. Then for any ɛ > 0, there is N, such that n > N implies l < ɛ. Also recall that, if we assume the sequence starts from n =, then the index of the subsequence satisfies n k k (prove this by induction!). Therefore k > N n k k > N k l < ɛ. The definition of it makes explicit use of the it value l. Therefore if we want to show the convergence of a sequence by the definition, we need to first know the value l and then find suitable N for each ɛ. In many cases, however, it is hard (or even impossible) to find the it of a convergent sequence. Proposition..7 gives an example where we know the convergence of a sequence without knowing the actual it value. The following criterion provides a method for the general case. Definition.. A sequence { } is called a Cauchy sequence if for any ɛ > 0, there is N, such that m, n > N implies x m < ɛ. Theorem.. (Cauchy Criterion) A sequence converges if and only if it is a Cauchy sequence. Theorem.. is called the Cauchy criterion for its of sequences, or just the Cauchy criterion for short. In fact, suppose { } converges to l. For any ɛ > 0, there is an N, such that n N l < ɛ.

8 Chapter. Limits and Continuity Then m, n > N x m l < ɛ, l < ɛ x m = ( l) (x m l) l + x m l < ɛ + ɛ = ɛ. The proof of the converse of the Cauchy criterion is much more difficult. It could be accomplished in the following three steps.. Any Cauchy sequence is bounded.. By the Bolzano-Weierstrass Theorem, any bounded sequence has a convergent subsequence. 3. If a Cauchy sequence has a subsequence converging to l, then the whole sequence converges to l. We leave the proof as an exercise to the readers. Example..36 In Example..8, we proved that the sequence { }, = + + 3 + + n. is increasing and bounded, so that it converges. Alternatively, we can prove that { } is a Cauchy sequence. In fact, for any ɛ > 0, n m > ɛ + x n x m = (m + ) + + n Hence { } coverges by the Cauchy criterion. (m )m + m(m + ) + + (n )n = ( m m ) + ( m m + ) + + ( n n ) = m n m < ɛ. Example..37 We prove that the sequence, called the harmonic series, = + + 3 + + n, diverges by showing that the Cauchy criterion fails. In fact, we show that there is ɛ > 0, such that for any N, we can find m, n > N satisfying x m ɛ. For any n, we have = n + + n + + + n n + n + + n =.

.. Limits of Sequences 9 Choose ɛ =. For any N, we may find a natural number n > N. Then we also have m = n > N and x m = = ɛ. Therefore, the sequence fails the Cauchy criterion and diverges. Exercises..9 Rigorously prove the its:. sin n + cos n = 0; n. n + sin n n cos n = ; 3. n + 3 n n = ; n! 4. n = 0; n 5. ( n + a n + b) = 0; 6. ( 3 n 3 n ) = 0; 7. n π n p 8. n! = 0...0 Prove that if = l and = k, then l = k... Prove that if = l, then = l... Prove that l = 0 if and only if = l...3 Prove that if = l, then (c ) = cl. = 0, a > ; an..4 Prove that if = l and = k 0, then = l y n k...5 Prove that a sequence { } converges if and only if the subsequences { } and {+ } converge to the same it...6 Prove that a monotonic sequence converges if and only if there is a convergent subsequence...7 Prove that if a bounded sequence { } diverges, then there are two subsequences that converge to different finite its...8 Prove Theorem.., the Cauchy criterion for its of sequences...9 If { } is a Cauchy sequence, can you conclude that { } is also a Cauchy sequence? Why? What about the converse?..30 Suppose the sequence { } satisfies Prove that { } is convergent...3 If does the sequence { } converge? Explain. + /n, n=,, 3,... = sin(π/n), n=,, 3,..., +..3 Suppose > 0 for all n. Prove that if = l, then n = l.

30 Chapter. Limits and Continuity..4 Infinite Limits A sequence may diverge for various reasons. For example, the sequence {n} diverges because it become arbitrarily large, so that the sequence is not bounded. For another example, the sequence {( ) n } is bounded but diverges because it has two subsequences with different finite its. The first case may be summarized by the following definition. Definition..3 A sequence diverges to infinity, denoted =, if for any B, there is N, such that n > N implies > B. In the definition, the infinity means that the absolute value (or the magnitude) of the sequence can get arbitrarily large. If we further take into account of the signs, then we get the definition of = + by requiring > B, and get the definition of = by requiring < B. For example, we have n = +. The meaning of = + is illustrated in Figure.5. x x 4 x N+ xn+ x N+3 B x x 3 N n Figure.5: n > N implies > B. Example..38 We have np = +, for p > 0. Here is the rigorous proof. For any B > 0, we have n > B p n p > B. Note that in the definition of = +, we may additionally assume B > 0 without loss of generality. Similarly, in the definition of =, we may also assume B < 0 (or B < 00 if needed). Example..39 The idea for Example..9 may also be used to show that a n = for a >. Specifically, let a = + b. Then a > implies b > 0, and we have a n = ( + b) n = + nb + n(n ) b + + b n > nb.

.. Limits of Sequences 3 For any B, choose N = B. Then we have b n > N a n > nb > Nb = B. This proves that a n =. If we take the sign into account, this also proves a n = + for a >. Example..40 Assume 0. We prove that = 0 implies =. Actually the converse is also true and the proof is left to the reader. If = 0, then for any B > 0, we apply the definition of the it to > 0. Thus B there is N, such that n > N < B > B. Therefore we conclude =. Applying what we just proved to the it in Example..9, we get another proof of an = for a >. A sequence { } is an infinitesimal if = 0. Example..40 shows that a nonzero sequence is an infinitesimal if and only if its reciprocal diverges to infinity. Many properties of the finite it can be extended. The following are some examples of the extended arithmetic rules. Example..40 is included in the first two rules. l 0 l = for l 0: If x n = l 0 and y n = 0, then =. y n = 0: If x n = l and y n =, then = 0. y n (+ ) + (+ ) = + : If = + and y n = +, then ( + y n ) = +. ( ) + l = : If = and y n = l, then ( + y n ) =. (+ ) l = for l < 0: If = + and y n = l < 0, then y n =. l = + for l > 0: If 0 + x n = l > 0, y n = 0 and y n > 0, then = +. y n The notations such as l = 0 are simply used as symbols for the underlying statements about its. One should also be cautious not to overly extend the arithmetic properties. For example, the following properties are actually wrong + =, + =, 0 = 0, 0 =.

3 Chapter. Limits and Continuity A counterexample for the first equality is = n and y n = n, for which we have =, y n = and ( + y n ) = 0. The following are the extensions of some other properties. Sandwich Rule: If y n and y n = +, then = +. Order Rule: If = + and y n = l is finite, then > y n for sufficiently large n. The following complements Proposition..7. Proposition..4 If a monotonic sequence { } is unbounded, then =. Specifically, an increasing sequence { } is bounded below by x. Hence only two exclusive cases can happen: { } is bounded above, or { } is not bounded above. If { } is bounded above, then Proposition..7 says that { } converges. If { } is not bounded above, then any number B is not an upper bound. In other words, there is some term a N > B. Then the sequence is increasing, we have n > N a N > B. This proves that { } diverges to +. Similar discussion can be made to decreasing sequences. If a decreasing sequence is bounded, then it converges. If the sequence is not bounded, then it diverges to. Example..4 By n = + and the extended arithmetic rule, we have n k = + for k. Then by the extended arithmetic rule again, we have (n3 3n + ) = n 3 ( 3 n + ) = (+ ) = +. n3 Example..4 For a >, the sequence {a n } is increasing. So either the sequence is bounded and converges to a finite it l, or the sequence is unbounded and diverges to +. If a n = l, then al= a a n = a n+ = l. Since the sequence is increasing, we have l a >. Therefore it is impossible to have al= l. We conclude that a n = +. Exercises..33 Prove the following results:

.. Limits of Functions 33. a n n =, for a > ;. n sin a =, for a 0, ±π, ±π,... ; 3. n 3n + n = + ; 4. n( n + n ) = +...34 Prove that if = + and p > 0, then x p n = +. What about the case p < 0?..35 Construct sequences { } and {y n }, such that both diverge to infinity, but { + y n } can have any of the following behaviors.. ( + y n ) =.. ( + y n ) =. 3. { + y n } is bounded but does not converge. 4. { + y n } is not bounded and does not diverge to infinity. This exercise shows that + has no definite meaning...36 Suppose = +. Show that the following arithmetic rules hold.. If {y n } is bounded below, then ( + y n ) = +.. If c > 0, then (c ) = +. 3. If y n > M for some M > 0, for sufficiently large n, then ( y n ) = +. y n 4. If {y n } is bounded and 0 for all n, then = 0.. Limits of Functions.. The Concept of Limit and Its Properties We have seen various behaviors that sequences may disply as the index goes to infinity. Similar observations can be made for functions near a point. For example, as x is approaching 0, the functions x and x also approach 0, x gets larger and does not approach a fixed finite number, and sin(x ) swings between and, and never approaches any one specific finite number. Among the three functions, only x and x converge as x goes to 0. We also note that, as we talk about the behavior as x approaches 0, the function does not have to be defined at 0. Analogously, as the index of a sequence goes to infinity, the index is never equal to infinity. Definition.. (Non-rigorous) If f(x) approaches a finite number l when x approaches (but not equal to) a, then we say the function f(x) converges to the it l at a and write f(x) = l. x a The function diverges at a if it does not approach a specific finite number when x approaches a.