1610912322-b551b095a53deaf3d3fbd1ed05ae9b84 (824701), страница 31
Текст из файла (страница 31)
The equivalence of these definitions for realvalued functions follows from the fact mentioned earlier that every neighborhood ofa point contains a symmetric neighborhood of the same point (carry out the proof infull!).We have now given the general definition of the limit of a function over a base.Earlier we considered examples of the bases most often used in analysis. In a specificproblem in which one or another of these bases arises, one must know how to decodethe general definition and write it in the form specific to that base.Thus,lim f (x) = A := ∀ε > 0 ∃δ > 0 ∀x ∈ ]a − δ, a[ f (x) − A < ε ,x→a−0lim f (x) = A := ∀ε > 0 ∃δ ∈ R ∀x < δ f (x) − A < ε .x→−∞12 Formore details, see Bourbaki’s, General topology, Addison-Wesley, 1966.3.2 The Limit of a Function129In our study of examples of bases we have in particular introduced the conceptof a neighborhood of infinity.
If we use that concept, then it makes sense to adoptthe following conventions in accordance with the general definition of limit:lim f (x) = ∞ := ∀V (∞) ∃B ∈ B f (B) ⊂ V (∞) ,Bor, what is the same,lim f (x) = ∞ := ∀ε > 0 ∃B ∈ B ∀x ∈ B ε < f (x) ,Blim f (x) = +∞ := ∀ε ∈ R ∃B ∈ B ∀x ∈ B ε < f (x) ,Blim f (x) = −∞ := ∀ε ∈ R ∃B ∈ B ∀x ∈ B f (x) < ε .BThe letter ε is usually assumed to represent a small number.
Such is not the casein the definitions just given, of course. In accordance with the usual conventions, forexample, we could writelim f (x) = −∞ := ∀ε ∈ R ∃δ ∈ R ∀x > δ f (x) < ε .x→+∞We advise the reader to write out independently the full definition of limit fordifferent bases in the cases of both finite (numerical) and infinite limits.In order to regard the theorems on limits that we proved for the special baseE x → a in Sect. 3.2.2 as having been proved in the general case of a limit over anarbitrary base, we need to make suitable definitions of what it means for a functionto be ultimately constant, ultimately bounded, and infinitesimal over a given base.Definition 13 A function f : X → R is ultimately constant over the base B if thereexists a number A ∈ R and an element B ∈ B such that f (x) = A for all x ∈ B.Definition 14 A function f : X → R is ultimately bounded over the base B if thereexists a number c > 0 and an element B ∈ B such that |f (x)| < c for all x ∈ B.Definition 15 A function f : X → R is infinitesimal over the base B if limB f (x)= 0.After these definitions and the fundamental remark that all proofs of the theoremson limits used only the properties B1 ) and B2 ), we may regard all the properties oflimits established in Sect.
3.2.2 as valid for limits over any base.In particular, we can now speak of the limit of a function as x → ∞ or as x →−∞ or as x → +∞.In addition, we have now assured ourselves that we can also apply the theory oflimits in the case when the functions are defined on sets that are not necessarily setsof numbers; this will turn out to be especially valuable later on. For example, the1303Limitslength of a curve is a numerical-valued function defined on a class of curves.
If weknow this function on broken lines, we can define it for more complicated curves,for example, for a circle, by passing to the limit.At present the main use we have for this observation and the concept of a baseintroduced in connection with it is that they free us from the verifications and formalproofs of theorems on limits for each specific type of limiting passage, or, in ourcurrent terminology, for each specific type of base.In order to master completely the concept of a limit over an arbitrary base, weshall carry out the proofs of the following properties of the limit of a function ingeneral form.3.2.4 Existence of the Limit of a Functiona. The Cauchy CriterionBefore stating the Cauchy criterion, we give the following useful definition.Definition 16 The oscillation of a function f : X → R on a set E ⊂ X isω(f, E) := sup f (x1 ) − f (x2 ),x1 ,x2 ∈Ethat is, the least upper bound of the absolute value of the difference of the values ofthe function at two arbitrary points x1 , x2 ∈ E.Example 11 ω(x 2 , [−1, 2]) = 4;Example 12 ω(x, [−1, 2]) = 3,Example 13 ω(x, ]−1, 2[) = 3;Example 14 ω(sgn x, [−1, 2]) = 2;Example 15 ω(sgn x, [0, 2]) = 1;Example 16 ω(sgn x, ]0, 2]) = 0.Theorem 4 (The Cauchy criterion for the existence of a limit of a function) Let Xbe a set and B a base in X.A function f : X → R has a limit over the base B if and only if for every ε > 0there exits B ∈ B such that the oscillation of f on B is less than ε.Thus,∃ lim f (x) ⇔ ∀ε > 0 ∃B ∈ B ω(f, B) < ε .B3.2 The Limit of a Function131Proof Necessity.
If limB f (x) = A ∈ R, then, for all ε > 0, there exists an elementB ∈ B such that |f (x) − A| < ε/3 for all x ∈ B. But then, for any x1 , x2 ∈ B wehave f (x1 ) − f (x2 ) ≤ f (x1 ) − A + f (x2 ) − A < 2ε ,3and therefore ω(f ; B) < ε.Sufficiency. We now prove the main part of the criterion, which asserts that if forevery ε > 0 there exists B ∈ B for which ω(f, B) < ε, then the function has a limitover B.Taking ε successively equal to 1, 1/2, . .
. , 1/n, . . . , we construct a sequenceB1 , B2 , . . . , Bn . . . of elements of B such that ω(f, Bn ) < 1/n, n ∈ N. Since Bn =∅, we can choose a point xn in each Bn . The sequence f (x1 ), f (x2 ), . . . , f (xn ), . . .is a Cauchy sequence. Indeed, Bn ∩ Bm = ∅, and, taking an auxiliary point x ∈Bn ∩ Bm , we find that |f (xn ) − f (xm )| < |f (xn ) − f (x)| + |f (x) − f (xm )| <1/n + 1/m. By the Cauchy criterion for convergence of a sequence, the sequence{f (xn ), n ∈ N} has a limit A.
It follows from the inequality established above,if we let m → ∞, that |f (xn ) − A| ≤ 1/n. We now conclude, taking accountof the inequality ω(f ; Bn ) < 1/n, that |f (x) − A| < ε at every point x ∈ Bn ifn > N = [2/ε] + 1.Remark This proof, as we shall see below, remains valid for functions with values inany so-called complete space Y . If Y = R, which is the case we are most interestedin just now, we can if we wish use the same idea as in the proof of the sufficiency ofthe Cauchy criterion for sequences.Proof Setting mB = infx∈B f (x) and MB = supx∈B f (x), and remarking thatmB1 ≤ mB1 ∩B2 ≤ MB1 ∩B2 ≤ MB2 for any elements B1 and B2 of the base B, wefind by the axiom of completeness that there exists a number A ∈ R separatingthe numerical sets {mB } and {MB }, where B ∈ B.
Since ω(f ; B) = MB − mB , wecan now conclude that, since ω(f ; B) < ε, we have |f (x) − A| < ε at every pointx ∈ B.Example 17 We shall show that when X = N and B is the base n → ∞, n ∈ N,the general Cauchy criterion just proved for the existence of the limit of a functioncoincides with the Cauchy criterion already studied for the existence of a limit of asequence.Indeed, an element of the base n → ∞, n ∈ N, is a set B = N ∩ U (∞) = {n ∈ N |N < n} consisting of the natural numbers n ∈ N larger than some number N ∈ R.Without loss of generality we may assume N ∈ N. The relation ω(f ; B) ≤ ε nowmeans that |f (n1 ) − f (n2 )| ≤ ε for all n1 , n2 > N .Thus, for a function f : N → R, the condition that for any ε > 0 there existsB ∈ B such that ω(f ; B) < ε is equivalent to the condition that the sequence {f (n)}be a Cauchy sequence.1323Limitsb. The Limit of a Composite FunctionTheorem 5 (The limit of a composite function) Let Y be a set, BY a base in Y , andg : Y → R a mapping having a limit over the base BY .
Let X be a set, BX a basein X and f : X → Y a mapping of X into Y such that for every element BY ∈ BYthere exists BX ∈ BX whose image f (BX ) is contained in BY .Under these hypotheses, the composition g ◦ f : X → R of the mappings f andg is defined and has a limit over the base BX and limBX (g ◦ f )(x) = limBY g(y).Proof The composite function g ◦ f : X → R is defined, since f (X) ⊂ Y . SupposelimBY g(y) = A.
We shall show that limBX (g ◦ f )(x) = A. Given a neighborhoodV (A) of A, we find BY ∈ BY such that g(BY ) ⊂ V (A). By hypothesis, there existsBX ∈ BX such that f (BX ) ⊂ BY . But then (g ◦ f )(BX ) = g(f (BX )) ⊂ g(BY ) ⊂V (A). We have thus verified that A is the limit of the function (g ◦ f ) : X → R overthe base BX .Example 18 Let us find the following limit:sin 7x=?x→0 7xlimIf we set g(y) =sin yyand f (x) = 7x, then (g ◦ f )(x) =sin 7x7x .In this case Y =limy→0 g(y) = limy→0 siny y= 1, we can apply the theoremR\0 and X = R. Sinceif we verify that for any element of the base y → 0 there is an element of the basex → 0 whose image under the mapping f (x) = 7x is contained in the given elementof the base y → 0.The elements of the base y → 0 are the deleted neighborhoods Ů Y (0) of thepoint 0 ∈ R.The elements of the base x → 0 are also deleted neighborhoods Ů X (0) of thepoint 0 ∈ R.
Let Ů Y (0) = {y ∈ R | α < y < β, y = 0} (where α, β ∈ R and α < 0,β > 0) be an arbitrary deleted neighborhood of 0 in Y . If we take Ů X (0) = {x ∈R | α7 < x < β7 , x = 0}, this deleted neighborhood of 0 in X has the property thatf (Ů X (0)) = Ů Y (0) ⊂ Ů Y (0).The hypotheses of the theorem are therefore satisfied, and we can now assert thatsin 7xsin y= lim= 1.x→0 7xy→0 ylimExample 19 The function g(y) = | sgn y|, as we have seen (see Example 3), has thelimit limy→0 | sgn y| = 1.The function y = f (x) = x sin x1 , which is defined for x = 0, also has the limitlimx→0 x sin x1 = 0 (see Example 1).However, the function (g ◦ f )(x) = | sgn(x sin x1 )| has no limit as x → 0.Indeed, in any deleted neighborhood of x = 0 there are zeros of the functionsin x1 , so that the function | sgn(x sin x1 )| assumes both the value 1 and the value 0 in3.2 The Limit of a Function133any such neighborhood.
By the Cauchy criterion, this function cannot have a limitas x → 0.But does this example not contradict Theorem 5?Check, as we did in the preceding example, to see whether the hypotheses of thetheorem are satisfied.Example 20 Let us show thatlimx→∞11+xx= e.Proof Let us make the following assumptions:Y = N,BY is the base n → ∞, n ∈ N;X = R+ = {x ∈ R | x > 0},f :X→YBX is the base x → +∞;fis the mapping x −→[x],where [x] is the integer part of x (that is, the largest integer not larger than x).Then for any BY = {n ∈ N | n > N} in the base n → ∞, n ∈ N there obviouslyexists an element BX = {x ∈ R | x > N + 1} of the base x → +∞ whose imageunder the mapping x → [x] is contained in BY .1 n) , and g2 (n) = (1 + n1 )n+1 , asThe functions g(n) = (1 + n1 )n , g1 (n) = (1 + n+1we know, have the number e as their limit in the base n → ∞, n ∈ N.By Theorem 4 on the limit of a composite function, we can now assert that thefunctions[x]1 [x]1(g ◦ f )(x) = 1 +,(g1 ◦ f ) = 1 +,[x][x] + 11 [x]+1(g2 ◦ f ) = 1 +[x]also have e as their limit over the base x → +∞.It now remains for us only to remark that1+1[x] + 1[x]1 x1 [x]+1< 1+< 1+x[x]for x ≥ 1.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
