MA1521_10S1_Ch01
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Chapter 1.Functions:Limits and Continuity§1.1LimitsIn this section we study the behaviour of functional values f (x )as x gets closer and closer to a .Let us look at some examples:Example.Let f :R →R be given by f (x )= x +2if x =22if x =2Describe the behaviour of f (x )as x tends to 2.Solution.From the graph of f(x),we see that f(x) approaches4when x gets closer and closer to2(but not reaching2itself).Thus,we say“the limit of f as x tends to2is equal to4.”We write:f(x)=4.limx→2Note that f(2)=2does not change the limit. ExampleLet D=R\{0}and consider the functionsin xf:D→R given by f(x)=Solution.Clearly if x=0,then sin0,whichis meaningless.Moreover0is notx=1.31 Example.Let f:R\{0}−→R,f(x)=Informal DefinitionLet f(x)be defined on an open interval I containing x0,except possibly at x0itself.If f(x)gets arbitrar-ily close to a number L when x is sufficiently close to x0,then we say that the limit of f(x)as x tends to x0is the number L and we writef(x)=L.limx→x0Remarkf(x)may not exist. We have seen that limit limx→x0Even if the limit exists,it is not affected by the value f(x0)(when this is defined).5Rules of LimitsIf limx→a f(x)=L and limx→ag(x)=L′,then the follow-ing statements are easy to verify: (1)limx→a(f±g)(x)=L±L′;(2)limx→a(fg)(x)=LL′;(3)limx→a fL′provided L′=0;(4)limx→akf(x)=kL for any real number k. RemarkFor rule(3),if L=0and L′=0,then limx→afg(x)may or may not exist.Give examples to convince yourself.6Example.Using the above rules,limits of anypolynomial p (x )=c 0+c 1x +c 2x 2+···+c m x mcan be evaluated easily.Indeed,lim x →a p (x )=c 0+c 1a +c 2a 2+···+c m am=p (a ).ExampleA rational function f (x )=p (x )lim x →a q (x )=p (a )Caution(1)If the limit of the denominator is zero,then wecannot apply rule(3)directly.In this case,the limit for the rational function may or may not exist.We shall see later how to deal with this.(2)It is wrong to assumex we saw that limx→af(x)can exist even if thefunction is not defined at x=a(and thus there is no such thing as f(a)).Polynomials and rational functions are very special!8Example.Find limx→2x2−4x+2=p(2)4=0.Example.Find limx→2x2−4§1.2One-sided LimitsIn calculating the limit of f(x)as x tends to a,some-times it is more convenient to consider two cases: (1)right-hand limit or right limit limx→a+f(x), where x→a and x>a;and(2)left-hand limit or left limit limx→a−f(x), where x→a and x<a.These are known as one-sided limits. Theoremlim x→a f(x)exists if and only if limx→a+f(x)and limx→a−f(x)both exist and are equalExample.The graph of the function f:[0,4]→R is given below:(1)limx→0+f(x)=1;(2)limx→1−f(x)=0and limx→1+f(x)=1;limx→1f(x)does not exist;(3)limx→2−f(x)=limx→2+f(x)=limx→2f(x)=1,but note that f(2)=2;(4)limx→3−f(x)=limx→3+f(x)=limx→3f(x)=2.It also happens thatf(3)=2;(5)limx→4−f(x)=1.(6)For every other a∈[0,4],limx→af(x)=f(a).11§1.3Limits Involving Infinity Sometimes,we would like to discuss the function values f(x)as the input value x increases withoutf(x)”,which is read“the bound.We write“limx→∞limit of f as x tends to infinity”.Similarly,as x decreases without bound,we can dis-cuss“limf(x)”,which is“the limit of f as x tends x→−∞to negative infinity”.Important RemarkNote that∞is notwrite“x=∞”.12Examples(Limits of rational functions as x→±∞.)Find(i)limx→∞−x3x2+5;(iii)limx→∞5x+2 7x+4=(ii)limx→−∞2x2−x+3x+33+53+0= 22x3−2=lim x→∞5x3x3=0+0§1.4ContinuityContinuity is an intuitive concept.Intuitively,a func-tion is continuous if we can draw its graph“in one stroke”,or“without lifting the pen from the paper”.Continuous at c Not continuous at cPrecise definition:A function f is continuous(a)at an interior point c of its domain iff(x)=f(c)limx→cf(x)=f(a) (b)at a left endpoint a of its domain if limx→a+f(x)=f(b).(c)at a right endpoint b of its domain if limx→b−14Remark.(The Continuity Test)According to the definition,to test whether a func-tion f is continuous at a point p,we need to do the following3things:(i)check that p is in the domain of f(that is,f(p)is defined);f(x)exists(or the appropriate one (ii)check that limx→psided limit if p is an end-point);and(iii)check that limf(x)(or the appropriate one sidedx→plimit)is equal to f(p).15Example.Recall an earlier example where the graph of the function f:[0,4]→R is given below:(1)f is continuous at0because f(0)is defined(f(0)=1),limf(x)=1,and these two valuesx→0+are equal.f(x)does (2)f is not continuous at1because limx→1not exist.f(x)=f(2).(3)f is not continuous at2because limx→2f(x)=2=f(3).(4)f is continuous at3because limx→3(5)f is continuous at4because limf(x)=1=f(4).x→4−16Examples(1)All polynomials p(x)are continuous on R becauselimx→ap(x)=p(a)for every a∈R(see page7). (2)Similarly,all rational functionsp(x)q(x)=p(a)pointsin D.171.5More Continuous Functions and Some PropertiesIn addition to polynomials and rational functions,it can be shown that the following functions are also continuous on their natural domainsx or x1/n,where n is a posi-tive integer)It is easy to verify that sums,differences,products, and quotients of continuous functions are contin-uous on their domains.18Example.Find the value of limx→3x−3x2+7−4.We cannot substitute x=3,and the numerator and denominator have no obvious common factors.The trick is to multiply the numerator and denominator by the conjugate expression√√x2+7+4)x2+7−4)(√x2+7+4)x2+7+4)x2+7+4√x2+7+43.19Intermediate Value Theorem.(IVT)A function f(x)which is continuous in a closed in-terval[a,b]takes on every value between f(a)and f(b).That is,if f(a)≤y0≤f(b),then there exists.c∈[a,b]such that f(c)=y0Remarks(1)This theorem may not hold if f is not continuous.(2)The following theorem is an immediate conse-quence of the IVT with y0=0:20TheoremLet f be a continuous function on the closed inter-val[a,b]and suppose f(a)<0<f(b).Then the equation f(x)=0has at least one solution in[a,b].Example.Show that f(x)=9x8+7x−8has a real root between0and1.Solution.Since the function f(x)is a polynomial, it follows that f(x)is continuous on R.In particular, f(x)is continuous on[0,1].Now,f(0)=−8<0and f(1)=8>0,i.e.f(0)<0<f(1).By IVT,there is a point x=c in[0,1]such that f(c)=0.Thus,c is a real root of the polynomial f(x)between0and1.2122Remarks.(1)The IVT only indicates a root in the interval[0,1];it does not provide an algorithm tofind the root.(2)In the above example,using some other mathe-matical software,the root is c≈0.836025. (3)As an exercise,use the IVT to show that there isanother root in[−2,0].23。
