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A Primer on Limits 1. Perhaps the simplest kind of limit is a sequential limit: a sequence b : N → R has a limit L provided each neighborhood of L contains a tail of b. The two terms neighborhood and tail need defining: (a) A neighborhood of a real number is any open interval containing the number, e.g., the open intervals (0, 2), (9/10, 101/100), (−5.1, 11.34), and (−∞, π) are each neighborhoods of the number 1. (b) A tail of a sequence b is any subsequence of b with the form: b(k), b(k + 1), b(k + 2), b(k + 3), . . . for some k ∈ N. In particular, b is a tail of itself. The three sequences 1/2, 1/3, 1/4, 1/5 . . . 1/10, 1/11, 1/12, 1/13 . . . 1/1000, 1/1001, 1/1002, 1/1003 . . . are each tails of b if b(n) = 1/n; that is, each are tails of the sequence: 1, 1/2, 1/3, 1/4, . . . 1/n, . . . . The definition gives precision to the intuitive idea of how a limit (when it exists) of a sequence should relate to its values. If a sequences has a limit it is said to converge (to its limit). Figure 1 is an attempt to illustrate a convergent sequence b with limit L. Figure 1: The open interval containing L also contains the tail: b(6), b(7), b(8), . . . b(1) b(3) • • b(5) • • b(4) ( . b(6) • ••• •• ••......•• • b(7) . . . L ) b(2) • Notation: Convergence of sequence b to its limit L can be denoted by lim b(n) = L, or lim b(n) = L, or limn b(n) = L, n→∞ n or lim b(n) = L, or even by b(n) −→ L. Example 1: If b(n) = 1/n then limn b(n) = 0 (in short; limn 1/n = 0). In general, it’s much to ask that a function has a formula describing its values (remember that a sequence is a function). In this case we have one: b(n) = 1/n. b is simple to write down: 1, 1/2, 1/3, 1/4, 1/5 . . . and has an obvious limit of 0 (what other limit could it have?). However, to verify that limn→∞ b(n) = 0 the above definition requires a demonstration that any neighborhood of 0 contains a tail of b. You should have a go at this... before reading the next paragraph. Here is a way your argument could go: If N is a neighborhood of 0, that is, if N is an open interval containing 0, then N contains an interval (0, c) for some positive number c (draw a simple picture to help you see this, e.g., c could be taken as the right endpoint of N). Whenever the index n is such that n > 1/c then 1/n < c. For such indices n, the values b(n) will belong to the interval (0, c) and, hence, to the neighborhood N. That is, N contains a tail of b. If you actually did try to prove limn→∞ b(n) = 0 using the definition but found it difficult to start an argument, don’t worry about it too much: it’s only because you’re not used to making these arguments. Like anything else, you can get better with practice. 2. Limits of the form lim f (x) = L: If f is a function defined in a neighborhood of a then limx→a f (x) = L provided to x→a each neighborhood N of L there corresponds a neighborhood O of a such that f (x) ∈ N whenever x ∈ O and x 6= a (cf., figure 2). The limit L may have nothing to do with the value of the function f at a. When it does, that is, if it happens that L = f (a) then f is said to be continuous at a. Figure 2 is a schematic of the expression limx→a f (x) = L. Figure 2: limt→a f (t) = L: A neighborhood O of a is found that is mapped by f into a given neighborhood N of L. f ......................................................... .................... ............. ............. ........... ........... ......... ......... ........ . . . . . . . ........ . ..... . . . . .......... . . . ....... ...................... O ......................... •y ............... a ............ .......... ......... x • ........ ........ z • ........ ..... ... ....... .............. ....... ....... ....... ....... ....... ....... ....... ........ ........................... ........ ........ ........ ......... ......... .......... ........... .............. ........................................... ........ N f (x)• • f (z) L f (y) ...........• ............... Example 2: A proof of the statement limx→1 (3x + 2) = 5. Suppose N is a neighborhood of 5. Then there is a positive number (perhaps small) such that the open interval (5 − , 5 + ) is subset of the neighborhood N. Now, for any number x, 1 |(3x + 2) − 5| = |3x − 3| = 3|x − 1|. Consequently, if a number x is such that |x − 1| < then 3 |(3x + 2) − 5| < . This is just another way of saying that if x is any number in the interval (1 − /3, 1 + /3) then that 3x + 2 belongs to the interval (5 − , 5 + ) and hence, N. Letting O denote the open interval (1 − /3, 1 + /3) finishes the proof. 3. Limits of the form limx→∞ f (x) = L: If f is a function whose domain contains an interval of the form (a, ∞) then limx→∞ f (x) = L provided to each neighborhood N of L there corresponds a neighborhood O of ∞ such that f (x) ∈ N whenever x ∈ O. (Neighborhood s of ∞ are R or any interval (ray) of the form (b, ∞).) Example 3: limx→∞ 1/x = 0. proof: If N is a neighborhood of 0 then N contains an interval of the form (0, b) for some positive number b. Now x > 1/b implies 0 < 1/x < b. So let O = (1/b, ∞). Then O is a neighborhood of ∞ and 1/x ∈ N whenever x ∈ O. Remark : An analogous definition holds for limx→−∞ f (x) = L, with neighborhoods of −∞ being R or rays (−∞, a), a ∈ R. Exercise 1: (a) A convergent sequence b has exactly one limit. (b) If limx→a f (x) converges then its limit is unique (a = ±∞ is allowed). 4. Sequential limits once again. If f is a function and x is a sequence then the composition f ◦ x (whose value at n is f (x(n))) is also a sequence, that is, f ◦ x : N → R. Now suppose that limx→a f (x) = L and x is any sequence such that limn→∞ x(n) = a. Then it’s a straightforward check, using definitions 1 and 2, that limn→∞ f (x(n)) = L. 2 Exercise 2: Check this. More precisely, argue the assumption limx→a f (x) = L implies that limn→∞ f (x(n)) = L whenever x is a sequence with limn→∞ x(n) = a. On the other hand, just knowing that limn f (x(n)) = L for some sequence x with limn→∞ x(n) = a is not enough to conclude limx→a f (x) = L. To see this take, for example, f to be the function f (x) = sin(1/x) for x 6= 0 (you can define f to be anything you want at 0) and consider the two sequences x(n) = 1/(nπ) and y(n) = 1/(π/2+2nπ) = 2/[(4n+1)π]. Both sequences x and y have limit 0 yet f (x(n)) = sin(nπ) = 0 for every n and f (y(n)) = sin[(4n + 1)nπ/2] = 1 for every n. This means that limn f (x(n)) = 0 and limn f (y(n)) = 1. So limx→0 f (x) cannot exist even though some sequential limits exist. However, if limn f (x(n)) = L for any sequence x with limn→∞ x(n) = a then it will follow that that limx→a f (x) = L. Here is a proof: if limx→a f (x) 6= L then there must exist a neighborhood N of L such that in each neighborhood (a − 1/n, a + 1/n) of a there is a point xn with f (xn ) not in N. Define a sequence x by x(n) = xn . Then limn x(n) = a yet limn f (x(n)), if it even exists, cannot be L, which contradicts assumption. The theorem below summarizes this result. Theorem 1. limx→a f (x) = L if and only if limn f (x(n)) = L for every sequence x with limn x(n) = a. Remark : Theorem 1 holds also with a = ∞. P∞ 5. Series: Corresponding to each sequence a is an infinite sum or series: a(1) + a(2) + a(3) + · · · = n=1 a(n). P∞ a(n) is associated not just the sequence a but another sequence S of finite or partial sums To each series n=1 S(n) = a(1) + a(2) + a(3) + · · · + a(n − 1) + a(n). P∞ The series n=1 a(n) is said to converge to (the sum) S provided its sequence S of partial sums converges to S, that is, provided limn→∞ S(n) = S. P Remark : Suppose n a(n) converges to S. Since S(n) = a(1) + a(2) + · · · + a(n − 1) + a(n) =P S(n − 1) + a(n) it follows that limn a(n) = limn [S(n) − S(n − 1)] = limn S(n) − limn S(n − 1) = S − S = 0. Thus, for n a(n) to converge it is necessary that limn a(n) = 0. However, to guarantee convergence P P∞ the condition limn a(n) = 0 is not in general sufficient P of n a(n). The harmonic series, n=1 1/n, provides an example of a divergent series n a(n) with limn a(n) = 0. 3