SinceM0, there exists a positive number d1such that for all x0xcd1⇒g xMM2.(3)continued
574AppendicesFor any numbers Aand Bit can be shown thatABABandBAAB,from which it follows thatABAB. With Ag xand BM, this becomesg xMg xM,which can be combined with the inequality on the right in (3) to get, in turn,g xMM2M2g xMM2M2g x32Mg1xM2g3x.Multiply by 2Mg x.(4)Therefore, 0xcdimplies thatg1xM1MMggxxM1•g1x•Mg xM1•M2•Mg x.Inequality (4)(5)Since 1 2M2e0, there exists a number d20 such that for all xin D0xcd2⇒Mg x2eM2.(6)If we take dto be the smaller of d1and d2, the conclusions in (5) and (6) both hold for all xsuch that 0xcd. Combining these conclusions gives0xcd2⇒g1xM1e.This completes the proof of the Limit Quotient Rule.■The last proof we give is of the Sandwich Theorem (Theorem 4) of Section 2.1.THEOREM 4The Sandwich TheoremIf g xf xh xfor all xcin some interval about c, andlimx→cg xlimx→ch xL,thenlimx→cf xL.Proof for Right-hand Limits Suppose that limx→cg xlimx→ch xL.Then for any e0 there exists a d0 such that for all xthe inequality cxcdimpliesLeg xLeandLeh xLe.continued
Section A3Using the Limit Definition575These inequalities combine with the inequality g xf xh xto giveLeg xf xh xLe,Lef xLe,ef xLe.Thus,for all x,the inequality cxcdimpliesf xLe.Therefore,limx→cf xL.Proof for Left-hand Limits Suppose that limx→cg xlimx→ch xL. Thenfor any e0 there exists a d0 such that for all xthe inequality cdxcimpliesLeg xLeandLeh xLe. We conclude as before that for all x,cdxcimpliesf xLe. Therefore,limx→cf xL.Proof for Two-sided Limits If limx→cg xlimx→ch xL, then g xandh xboth approach Las x→cand x→c; so limx→cf xLand limx→cf xL.Hence limx→cf xexists and equals L.■Section A3 ExercisesIn Exercises 1 and 2, sketch the interval a,bon the x-axis with thepoint cinside. Then find a value of d0 such that for all x,0xcd⇒axb.1.a4 9,b4 7,c1 22.a2.7591,b3.2391,c3In Exercises 3 and 4, use the graph to find a d0 such that for all x 0xcd⇒f xLe.3.4.Exercises 5–8 give a functionf xand numbers L,c, and e.Find an open interval about con which the inequalityf xLeholds. Then give a value for d0 such that for all xsatisfying 0xcdthe inequalityf xLeholds. Use algebra to find your answers.5.f x2x2,L6,c2,e0.026.f xx1,L1,c0,e0.17.f x19x,L3,c10,e18.f xx2,L4,c2,e0.5Exercises 9–12 give a functionf x, a point c, and a positive number e.(a)Find Llimx→cf x. Then (b)find a number d0such that forall x0xcd⇒f xLe.