THE TEACHING OF MATHEMATICS 29, Vol. XII, 1, pp. 1 6 LIMITS OF COMPOSITE FUNCTIONS Milosav M. Marjanović and Zoran Kadelburg Abstract. Finding gf)), first the following two its i) f) = y ii) gy) = α are found and then, it is taken that gf)) = α. The eistence of the its under i) and ii) is the basis for this method, which is not legitimate in general. In this notice we give necessary and sufficient conditions for the legitimacy of this method. ZDM Subject Classification: I25; AMS Subject Classification: A35. Key words and phrases: Composite function; introduction of a new variable. 1. The it of a composite function g f at a point is often found by finding i) f) = y ii) gy) = α and then, by taking that gf)) = α. This procedure certainly establishes a method of finding the it of a composite function. In some books on analysis this method is applied without any eplicit considerations of its legitimacy, in some others sufficient conditions of this or that form are given see, e.g., [1]) and, as far as we can follow it, the book [2] was the first place where necessary and sufficient conditions for the applicability of this method were given. In the cited book, the it of a function at a point was taken as the possibility of its continuous etension to that point. Also the cases of isolated points were included. That all made the corresponding proofs somewhat complicated. In this notice we omit the trivial) cases of isolated points and provide straighter proofs based on the Heine s definition of it. Let us note that setting of our considerations in the frame of metric spaces makes no difference when compared with the very specific case of real functions whose domains and ranges are subsets of R). Let us also add that for a set A, A will be the set of its accumulation points. 2. As a definition of the it we take the following Cauchy s version. Definition. Let M 1, d 1 ) and M 2, d 2 ) be metric spaces, A M 1, f : A M 2 and A. An element y M 2 is said to be the it of f at the point if C) for each real number ε > there eists a real number δ > such that A \ { } and d 1, ) < δ imply d 2 f), y ) < ε. The notations f) = y or f = y are used to denote that y is the it of f at.
2 M. M. Marjanović, Z. Kadelburg It is easy to show that the Cauchy s condition C) is equivalent to the following Heine s condition: H) for each sequence n ) in A \ { }, when n in M 1 ), then f n ) y in M 2 ). Indeed, let the condition C) hold true and let n ) be a sequence in A \ { } such that n. Then, for the given ε >, there eists δ > such that A \ { } and d 1, ) < δ imply d 2 f), y ) < ε. Since n, there eists a natural number n such that d 1 n, ) < δ for each n > n. It follows that, for n > n, d 2 f n ), y ) < ε, what proves that f n ) y. Therefore, the condition H) holds true. To prove the converse, suppose that the condition C) is not satisfied. Then, there eists a number ε > such that for each δ >, there eists A \ { } such that d 1, ) < δ and d 2 f), y ) ε. In particular, we can take δ = 1/n for each natural n and then find n A \ { } such that d 1 n, ) < 1/n and d 2 f n ), y ) ε. Hence, n and f n ) y, what means that the Heine s condition is not satisfied, either. Thus, the equivalence of conditions C) and H) has been proved. The following eample shows that the described method of finding the it of a composite function does not work without additional requirements. Eample 1. Let f : R R and g : R R be the functions defined by { 1,, f) =, for each R and g) =, =. Then, obviously, f) = and gy) = 1, but g f) = for each R and thereby g f) =. Thus, we have f) = y and gy) = α, but g f) α. The theorem that follows gives the necessary and sufficient conditions under which the method is legitimate. Theorem 1. Let M 1, M 2, M 3 be metric spaces, A M 1, B M 2 and let A, y B. Further on, let f : A B and g : B M 3 be functions such that Then I) f) = y and II) gy) = α M 3. g f) = α if and only if one of the following conditions holds: 1) gy ) = α, 2) there eists δ > such that for each A \ { }, d 1, ) < δ implies f) y.
Limits of composite functions 3 Proof. Suppose first that condition 1) holds. Let n ) be a sequence in A \ { } such that n. Assumption I) implies that f n ) y. Consider the following two subsequences of the sequence f n )) the first is formed by those terms y n ) which are different from y and the second by those terms z n ) which are equal to y. One of these subsequences might be finite; and then it can be neglected.) Now, y n B \ {y } and y n y, so, applying assumption II), we obtain that gy n ) α. For the subsequence z n ), condition 1) implies that gz n ) α. Hence, both of the possible subsequences of g f n )) tend to α, and so g f n ) α. Hence, we conclude that g f) = α. Suppose now that condition 2) holds and let us fi the corresponding real number δ. Let again n ) be a sequence in A \ { } such that n. Then, there eists n N such that for n > n, d 1 n, ) < δ, and so, by 2), f n ) y. It means that f n ) B \ {y } possibly with a finite number of eceptions), and we can apply assumption II) to conclude that g f n ) α. It follows that g f) = α is valid in this case, as well. Finally, we have to show that validity of one of the conditions 1) and 2) is necessary for the conclusion g f) = α. Suppose the contrary, that neither of them is valid. Let us notice first that y B, for otherwise the condition 2) would be satisfied. Then gy ) α and for each δ > there eists A \ { } such that d 1, ) < δ and f) = y. Take, for each n N, δ n = 1/n, and find the corresponding n A\{ } such that d 1 n, ) < 1/n and f n ) = y. Then n ) is a sequence in A \ { } such that n. But, g f n )) is a constant sequence with terms equal to gy ), hence different from α. It means that g f n ) α and thereby g f) = α does not hold. This completes the proof of the theorem. Note that neither of conditions 1) and 2) was fulfilled in Eample 1. A standard procedure of applying Theorem 1 for finding its can be formulated in the following way. Corollary 1. Let f : A B be an invertible function and f 1 y) = ; further on, let g f) = α. Then gy) = α. Proof. Indeed, I) M 2 B f 1 f M 1 A g M 3 f 1 y) =, II) g f) = α and the condition 2) there eists ε > such that y B \ {y } and d 2 y, y ) < ε imply f 1 y) is fulfilled because of the invertibility of f). Hence, by Theorem 1, we have α = g f)f 1 y)) = gy). In this form, the method is usually called introduction of a new variable.
4 M. M. Marjanović, Z. Kadelburg Before considering further eamples, note that when we deal with real functions f : A R A R), points ± can be included in consideration as elements of a metric space R, d). Here R = R {, } and a possible choice of metric d on R is given in Eercise 3 at the end of the paper. Neighbourhoods of the point have the form of open intervals M, ), and similarly for the point. In particular, a sequence n ) in R converges to α R if and only if the function : N R defined by n) = n has the it α in the metric space R, d). Eample 2. a) 1 + 1 ) = e; b) 1 + 1 ) = e; c) 1 + y) 1/y = e. Proof. a) Let B = N and g 1, g 2 : B R be given by g 1 n) = 1 + 1 ) n, g 2 n) = 1 + 1 n+1. n + 1 n) It follows easily from the definition of the number e that g 1 n) = g 2 n) = e. Let A = { R 1}, and f : A N be given as f) =. Obviously, f) =. Condition 2) of Theorem 1 is fulfilled for each A ), f). Therefore the functions ) 1 g 1 f)) = 1 + and g 2 f)) = 1 + 1 ) +1 + 1 have the it equal to e at the point. Since for each > 1 we have ) 1 1 + < 1 + 1 < 1 + + 1 ) 1 ) +1, the conclusion a) follows from the Sandwich Theorem. b) Using the substitution t = we obtain that 1 + 1 = 1 ) 1 ) t = 1 + 1 ) t. t t 1 Now the substitution u = t 1 gives 1 + 1 ) t = t 1 1 + 1 u) u+1 = 1 + u) 1 u 1 + 1 ) = e. u Both of these introductions of new variables can be easily justified. conclusion b) follows. Hence, the c) Let A = B = { R > } and f : A B be given by f) = 1/. Then f is invertible, f 1 y) = 1/y and f 1 y) =. Let g : B R be given + by gy) = 1 + y) 1/y. Then g f : B R acts like g f) = 1 + ) 1 and, by a), g f) = e. Applying Corollary 1 we obtain that 1 + + y)1/y = e.
Limits of composite functions 5 In a similar way, using b), it can be proved that 1 + y) 1/y = e, hence the conclusion c) follows. Eample 3. Let f : 1, ), ) be given by f) = 1 + ) 1/ and g :, ) R by gy) = log a y. Then, according to the previous eample, f) = e = y ), e gy) = log a e = α). Moreover, condition 1) of Theorem 1 is fulfilled gy ) = ge) = log a e = α. Therefore log a 1 + ) = log a 1 + ) 1/ = log a e. In particular, for ln = log e, we have ln1 + ) = 1. a y 1 Eample 4. In order to find, put y = log y a 1 + ) = f). Then f is an invertible function with f 1 y) = a y 1, and f 1 y) =. Furthermore, according to Eample 3, g f) = alog a 1+) 1 log a 1 + ) = log a 1 + ) Applying Corollary 1, we obtain that gy) = a y 1 y Since and g f) = 1 = ln a. log a e = ln a. We conclude with an eample involving functions in two variables. Eample 5. Find 1 + 1 ) 2 y +y, for a R.,a) We can write the given epression as,a) 1 + 1 ) 2 y y +y = e +y ln ) 1 + 1. y = a and using the continuity of logarithmic and eponential + y ln y = ln y for y > and e y = e for y R), we functions i.e., that obtain that the given it is equal to e a. Eercises 1. Let M 1 and M 2 be metric spaces, A M 1, B M 2 and f : A R, g : B R. Let f = α, g = β and let h: R 2 R be a continuous function. Then hf), gy)) = hα, β).,y )
6 M. M. Marjanović, Z. Kadelburg In particular, 2. a) b),,...,), ),y ),y ) e 1 2... n 1 = 1. 1 2... n n m sin 1 ) = 1. nm [f) + gy)] = [f) gy)] = f) + gy), f) gy). 3. The function ϕ: R 1, 1) given by ϕ) = 2 arctan is bijective and both π functions ϕ and ϕ 1 are continuous. a) The function d: R R [, ) given by d, y) = ϕ) ϕy) is a metric on R. b) A sequence n ) in R converges to α R if and only if the function n) = n has the it α in the metric space R, d). c) Let ϕ ) = 1 and ϕ) = 1. Then d, y) = ϕ) ϕy) defines a metric on R and R, d) is a subspace of R, d). d) Let f : A M. Then f, as it is usually defined, eists, where A R, if and only if f eists, where A is taken to be a subset of R, d), and these two its are equal. e) Let A M. Then f =, as it is usually defined, where f : A R, if and only if f =, where f : A R, with the metric d. REFERENCES 1. E. Landau, Einführung in die Differentialrechnung und Integralrechnung, 1934. 2. M. Marjanović, Mathematical Analysis, I in Serbian), Naučna knjiga, Beograd, 1979. Milosav M. Marjanović, SANU, Knez Mihailova 35, 11 Belgrade, Serbia E-mail: milomar@beotel.rs Zoran Kadelburg, Faculty of Mathematics, Studentski trg 16, 11 Belgrade, Serbia E-mail: kadelbur@matf.bg.ac.rs