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Then in Ů E (a)121we shall also have |g(x)|< |B|, that is, the function g(x)is ultimately bounded asE x → a. We then write Af (x) A A + α(x) Af(x) − =− =− =gBg(x) BB + β(x) B=11·Bα(x) + Aβ(x) = γ (x).g(x) B10 Here is a curious detail. This very obvious representation, which is nevertheless very usefulon the computational level, was specially noted by the French mathematician and specialist inmechanics Lazare Carnot (1753–1823), a revolutionary general and academician, the father ofSadi Carnot (1796–1832), who in turn was the creator of thermodynamics.3.2 The Limit of a Function115By the properties of infinitesimals (taking account of the ultimate boundedness of1g(x) ) we find that the function γ (x) is infinitesimal as E x → a.
Thus we haveproved that limEx→a ( fg )(x) =AB.c. Passage to the Limit and InequalitiesTheorem 3 a) If the functions f : E → R and g : E → R are such thatlimEx→a f (x) = A, and limEx→a g(x) = B and A < B, then there exists adeleted neighborhood Ů E (a) of a in E at each point of which f (x) < g(x).b) If the relations f (x) ≤ g(x) ≤ h(x) hold for the functions f : E → R, g :E → R, and h : E → R, and if limEx→a f (x) = limEx→a h(x) = C, then thelimit of g(x) exists as E x → a, and limEx→α g(x) = C.Proof a) Choose a number C such that A < C < B.
By definition of limit, we finddeleted neighborhoods Ů E (a) and Ů E (a) of a in E such that |f (x) − A| < C − Afor x ∈ Ů E (a) and |g(x) − B| < B − C for x ∈ Ů E (a). Then at any point of adeleted neighborhood Ů E (a) contained in Ů E (a) ∩ Ů E (a), we findf (x) < A + (C − A) = C = B − (B − C) < g(x).b) If limEx→a f (x) = limEx→a h(x) = C, then for any fixed ε > 0 there existdeleted neighborhoods Ů E (a) and Ů E (a) of a in E such that C − ε < f (x) forx ∈ Ů E (a) and h(x) < C + ε for x ∈ Ů E (a). Then at any point of a deleted neighborhood Ů E (a) contained in Ů E (a) ∩ Ů E (a), we have C − ε < f (x) ≤ g(x) ≤h(x) < C + ε, that is, |g(x) − C| < ε, and consequently limEx→a g(x) = C.
Corollary Suppose limEx→a f (x) = A and limEx→a g(x) = B. Let Ů E (a) be adeleted neighborhood of a in E.a)b)c)d)If f (x) > g(x) for all x ∈ Ů E (a), then A ≥ B;f (x) ≥ g(x) for all x ∈ Ů E (a), then A ≥ B;f (x) > B for all x ∈ Ů E (a), then A ≥ B;f (x) ≥ B for all x ∈ Ů E (a), then A ≥ B.Proof Using proof by contradiction, we immediately obtain assertions a) and b) ofthe corollary from assertion a) of Theorem 3. Assertions c) and d) follow from a)and b) by taking g(x) ≡ B.d. Two Important ExamplesBefore developing the theory of the limit of a function further, we shall illustrate theuse of the theorems just proved by two important examples.1163LimitsFig.
3.1Example 9sin x= 1.x→0 xlimHere we shall appeal to the definition of sin x given in high school, that is, sin x isthe ordinate of the point to which the point (1, 0) moves under a rotation of x radiansabout the origin. The completeness of such a definition is entirely a matter of thecare with which the connection between rotations and real numbers is established.Since the system of real numbers itself was not described in sufficient detail in highschool, one may consider that we need to sharpen the definition of sin x (and thesame is true of cos x).We shall do so at the appropriate time and justify the reasoning that for now willrely on intuition.a) We shall show thatcos2 x <πsin x< 1 for 0 < |x| < .x2Proof Since cos2 x and sinx x are even functions, it suffices to consider the case 0 <x < π/2.
By Fig. 3.1 and the definition of cos x and sin x, comparing the area of thesector OCD, the triangle OAB, and the sector OAB, we have111SOCD = |OC| · |CD| = (cos x)(x cos x) = x cos2 x <222111< S OAB = |OA| · |BC| = · 1 · sin x = sin x <222111< SOAB = |OA| · |AB| = · 1 · x = x.222Dividing these inequalities by 12 x, we find that the result is what was asserted. b) It follows from a) that| sin x| ≤ |x|for any x ∈ R, equality holding only at x = 0.3.2 The Limit of a Function117Proof For 0 < |x| < π/2, as shown in a), we have| sin x| < |x|.But | sin x| ≤ 1, so that this last inequality also holds for |x| ≥ π/2 > 1. Only forx = 0 do we find sin x = x = 0.c) It follows from b) thatlim sin x = 0.x→0Proof Since 0 ≤ | sin x| ≤ |x| and limx→0 |x| = 0, we find by the theorem on thelimit of a function and inequalities (Theorem 3) that limx→0 | sin x| = 0, so thatlimx→0 sin x = 0.d) We shall now prove that limx→0sin xx= 1.Proof Assuming that |x| < π/2, from the inequality in a) we have1 − sin2 x <sin x< 1.xBut limx→0 (1 − sin2 x) = 1 − limx→0 sin x · limx→0 sin x = 1 − 0 = 1, so thatby the theorem on passage to the limit and inequalities, we conclude thatlimx→0 sinx x = 1.Example 10 (Definition of the exponential, logarithmic, and power functions usinglimits) We shall now illustrate how the high-school definition of the exponentialand logarithmic functions can be completed by means of the theory of real numbersand limits.For convenience in reference and to give a complete picture, we shall start fromthe beginning.a) The exponential function.
Let a > 1.10 For n ∈ N we define inductively a 1 := a, a n+1 := a n · a.In this way we obtain a function a n defined on N, which, as can be seen from thedefinition, has the propertyam= a m−nanif m, n ∈ N and m > n.20 This property leads to the natural definitionsa 0 := 1,a −n :=1anfor n ∈ N,1183Limitswhich, when carried out, extend the function a n to the set Z of all integers, and thena m · a n = a m+nfor any m, n ∈ Z.30 In the theory of real numbers we have observed that for a > 0 and n ∈ N thereexists a unique nth root of a, that is, a number x > 0 such that x n = a. For thatnumber we use the notation a 1/n . It is convenient, since it allows us to retain the lawof addition for exponents:na = a 1 = a 1/n = a 1/n · · · a 1/n = a 1/n+···+1/n .For the same reason it is natural to set a m/n := (a 1/n )m and a −1/n := (a 1/n )−1for n ∈ N and m ∈ Z.
If it turns out that a (mk)/(nk) = a m/n for k ∈ Z, we can considerthat we have defined a r for r ∈ Q.40 For numbers 0 < x, 0 < y, we verify by induction that for n ∈ N(x < y) ⇔ x n < y n ,so that, in particular,(x = y) ⇔ x n = y n .50 This makes it possible to prove the rules for operating with rational exponents,in particular, thata (mk)/(nk) = a m/nfor k ∈ Zanda m1 /n1 · a m2 /n2 = a m1 /n1 +m2 /n2 .Proof Indeed, a (mk)/(nk) > 0 and a m/n > 0. Further, since (mk)/(nk) nk 1/(nk) mk nka= a=mk·nk 1/(nk) nk mk= a 1/(nk)= a= a mkand m/n nk 1/n n mka= a= a mk ,it follows that the first of the inequalities that needed to be verified in connectionwith point 40 is now established.Similarly, sincen nn n n n m /na 1 1 · a m2 /n2 1 2 = a m1 /n1 1 2 · a m2 /n2 1 2 =n m n n m n= a 1/n1 1 1 2 · a 1/n2 2 2 1 = a m1 n2 · a m2 n1= a m1 n2 +m2 n13.2 The Limit of a Function119and m /n +m /n n1 n2 (m n +m n )/(n n ) n1 n2= a 1 2 2 1 1 2=a 1 1 2 2 1/(n n ) n1 n2 m1 n2 +m2 n1= a m1 n2 +m2 n1 ,= a 1 2the second equality is also proved.Thus we have defined a r for r ∈ Q and a r > 0; and for any r1 , r2 ∈ Q,a r1 .
a r2 = a r1 +r2 .60 It follows from 40 that for r1 , r2 ∈ Q(r1 < r2 ) ⇒ a r1 < a r2 .Proof Since (1 < a) ⇔ (1 < a 1/n ) for n ∈ N, which follows immediately from 40 ,we have (a 1/n )m = a m/n > 1 for n, m ∈ N, as again follows from 40 . Thus for 1 < aand r > 0, r ∈ Q, we have a r > 1.Then for r1 < r2 we obtain by 50a r2 = a r1 · a r2 −r1 > a r1 · 1 = a r1 .70 We shall show that for r0 ∈ Qlim a r = a r0 .Qr→r0Proof We shall verify that a p → 1 as Q p → 0. This follows from the fact thatfor |p| < n1 we have by 60a −1/n < a p < a 1/n .We know that a 1/n → 1 (and a −1/n → 1) as n → ∞.
Then by standard reasoningwe verify that for ε > 0 there exists δ > 0 such that for |p| < δ we have1 − ε < a p < 1 + ε.We can take n1 as δ here if 1 − ε < a −1/n and a 1/n < 1 + ε.We now prove the main assertion.Given ε > 0, we choose δ so that1 − εa −r0 < a p < 1 + εa −r0for |p| < δ.
If now |r − r0 | < δ, we havea r0 1 − εa −r0 < a r = a r0 · a r−r0 < a r0 1 + εa −r0 ,which saysa r0 − ε < a r < a r0 + ε.1203LimitsThus we have defined a function a r on Q having the following properties:a 1 = a > 1;a r1 · a r2 = a r1 +r2 ;a r1 < a r2for r1 < r2 ;a →aas Q r1 → r2 .r1r2We now extend this function to the entire real line as follows.80 Let x ∈ R, s = supQr<x a r , and i = infQr>x a r .
It is clear that s, i ∈ R, sincefor r1 < x < r2 we have a r1 < a r2 .We shall show that actually s = i (and then we shall denote this common valueby a x ).Proof By definition of s and i we havea r1 ≤ s ≤ i ≤ a r2for r1 < x < r2 .
Then 0 ≤ i − s ≤ a r2 − a r1 = a r1 (a r2 −r1 − 1) < s(a r2 −r1 − 1). Buta p → 1 as Q p → 0, so that for any ε > 0 there exists δ > 0 such that a r2 −r1 − 1 <ε/s for 0 < r2 − r1 < δ. We then find that 0 ≤ i − s ≤ ε, and since ε > 0 is arbitrary,we conclude that i = s.We now define a x := s = i.90 Let us show that a x = limQr→x a r .Proof Taking 80 into account, for ε > 0 we find r < x such that s − ε < a r ≤ s =a x and r such that a x = i ≤ a r < i + ε.
Since r < r < r implies a r < a r < a r ,we then have, for all r ∈ Q in the open interval ]r , r [,a x − ε < a r < a x + ε.We now study the properties of the function a x so defined on R.100 For x1 , x2 ∈ R and a > 1, (x1 < x2 ) ⇒ (a x1 < a x2 ).Proof On the open interval ]x1 , x2 [ there exist two rational numbers r1 < r2 . Ifx1 ≤ r1 < r2 ≤ x2 , by the definition of a x given in 80 and the properties of thefunction a x on Q, we havea x1 ≤ a r1 < a r2 ≤ a x2 .110 For any x1 , x2 ∈ R, a x1 · a x2 = a x1 +x2 .Proof By the estimates that we know for the absolute error in the product and byproperty 90 , we can assert that for any ε > 0 there exists δ > 0 such thata x1 · a x2 −εε< a r1 · a r2 < a x1 · a x2 +223.2 The Limit of a Function121for |x1 − r1 | < δ , and |x2 − r2 | < δ . Making δ smaller if necessary, we can chooseδ < δ such that we also havea r1 +r2 −εε< a x1 +x2 < a r1 +r2 +22for |x1 − r1 | < δ and |x2 − r2 | < δ, that is, |(x1 + x2 ) − (r1 + r2 )| < 2δ.But a r1 · a r2 = a r1 +r2 , for r1 , r2 ∈ Q, so that these inequalities implya x1 · a x2 − ε < a x1 +x2 < a x1 · a x2 + ε.Since ε > 0 is arbitrary, we conclude thata x1 · a x2 = a x1 +x2 .120 limx→x0 a x = a x0 .
(We recall that “x → x0 ” is an abbreviation of “R x →x0 ”.)Proof We first verify that limx→0 a x = 1. Given ε > 0, we find n ∈ N such that1 − ε < a −1/n < a 1/n < 1 + ε.Then by 100 , for |x| < 1/n we have1 − ε < a −1/n < a x < a 1/n < 1 + ε,that is, we have verified that limx→0 a x = 1.If we now take δ > 0 so that |a x−x0 − 1| < εa −x0 for |x − x0 | < δ, we finda x0 − ε < a x = a x0 a x−x0 − 1 < a x0 + ε,which verifies that limx→x0 a x = a x0 .130 We shall show that the range of values of the function x → a x is the set R+of positive real numbers.Proof Let y0 ∈ R+ . If a > 1, then as we know, there exists n ∈ N such that a −n <y0 < a n .By virtue of this fact, the two setsand B = x ∈ R | y0 < a xA = x ∈ R | a x < y0are both nonempty.
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