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Lecture 5 : Continuous Functions
Denition 1We say the functionfis continuous at a numberaif lim x!af(x) =f(a): (i.e. we can make the value off(x) as close as we like tof(a) by takingxsuciently close toa). ExampleLast day we saw that iff(x) is a polynomial, thenfis continuous atafor any real number asince limx!af(x) =f(a). Iffis dened for all of the points in some interval arounda(includinga), the denition of continuitymeans that the graph is continuous in the usual sense of the word, in that we can draw the graph as a
continuous line, without lifting our pen from the page. Note that this denition implies that the functionfhas the following three propertiesiffis continuous ata:1.f(a) is dened (ais in the domain off).
2. li m x!af(x) exists. 3. li m x!af(x) =f(a). (Note that this implies that limx!af(x) and limx!a+f(x) both exist and are equal).Types of DiscontinuitiesIf a functionfis dened neara(fis dened on an open interval containinga, except possibly ata),
we say thatfis discontinuousata(or has a discontinuiuty ata) iffis not continuous ata. This can happen in a number of ways. In the graph below, we have a catalogue of discontinuities. Note that a function is discontinuous ataif at least one of the properties 1-3 above breaks down. Example 2Consider the graph shown below of the function k(x) =8 >>>:x23< x <3
x3x <5 0x= 5 x5< x71x10x >7
Where is the function discontinuous and why?1
We sayf(x) has a removable discontinuityataif we can remove the discontinuity ata, by changing the value of the function ata(makingf(a) = limx!af(x)). Which of the above discontinuities are removable discontinuities? Discontinuities where the graph has a vertical asymptote are called innite discontinuities. Which of the discontinuities above are innite discontinuities?Those discontinuities where the graph jumps are called jumpdiscontinuities. Which of the discontinuities
above are jump discontinuities? DenitionA functionfis continuous from the right at a number aif lim x!a+=f(a). A functionfis continuous from the left at a number aif lim x!a=f(a). Example 3Consider the functionk(x) in example 2 above. Ar which of the following x-values is k(x) continuous from the right? x= 0; x= 3; x= 5; x= 7; x= 10: At which of the above x-values isk(x) continuous from the left? DenitionA functionfis continuous on an intervalif it is continuous at every number in the interval. (Iffis dened only on one side of an endpoint of the interval, we understandcontinuousat the endpoint to meancontinuous from the right or continuous from the leftat the endpoint as appropriate. ) ExampleConsider the functionk(x) in example 2 above. (a) On which of the following intervals isk(x) continuous? (1;0];(1;3);[3;7]: (b) Fill in the missing endpoints and brackets which give the largest intervals on whichk(x) is continuous. (1;(5; 2ExampleLet
m(x) =cx2+ 1x210x x <2
For which value ofcism(x) a continuous function?
Catalogue of functions continuous on their domainsFrom the last day we know:
Polynomials and Rational functions
A polynomial function,P(x) =c0+c1x+c2x2++cnxn, is continuous everywhere i.e. lim x!aP(x) =P(a) for all real numbersa. A rational function,f(x) =P(x)Q(x), whereP(x) andQ(x) are polynomials is continuous on its domain, i.e. lim x!af(x) =P(a)Q(a)for all values ofa, whereQ(a)6= 0. n th Root function From #10 in last day's lecture, we also have that iff(x) =npx, wherenis a positive integer, thenf(x) is continuous on the interval [0;1). We can use symmetry of graphs to extend this to show thatf(x) is continuous on the interval (1;1), whennis odd. Hence all n th root functions are continuous on their domains.Trigonometric Functions
In the appendix we provide a proof of the following Theorem :Theorem 1The functions sinxand cosxare continuous on the interval (1;1). In particular;
for any real numbera, we can evaluate the limits below by direct substitution limx!asinx= sina;limx!acosx= cosa:Combining this with Theorem 2 below will show that all of the trigonometric function sinx;cosx,
tanx;secx, cscx;cotxare continuous on their domains. From Theorem 2 below we get thatfunctions which are algebraic combinations of the functions using+;;andlisted above are also continuous on their domains.Theorem 2Iffandgare continuous ataandcis constant, then the following functions are
also continuous ata:1: f+g2: fg3: cf4: fg5:fg
ifg(a)6= 0:3It is relative easy to prove this theorem using the limit laws from the previous lecture. A sample proof
of number 5 is given in the appendix.NoteCollecting the above results, we can show that the following types of functions and
combinations of them using +;;and are continuous on their domains:Polynomial Functions Rational functions
Root Functions Trigonometric functionsExampleFind the domain of the following function and use the theorem above to show that it is
continuous on its domain:g(x) =(x2+3)2x10: ExampleFind the domain of the following function and use the theorem above to show that it is continuous on its domain: k(x) =3px(x2+ 2x+ 1) +x+ 1x10:k(x) is continuous on its domain, since it is a combination of root functions, polynomials and rational
functions using the operations +;;and . The domain ofkis all values ofxexceptx= 10 and this function is continuous on the intervals (1;10) and (10;1). ExampleUse Theorem 1 and property 5 from theorem 2 above to show that tanxis continuous on its domain. tan(x) is continuous on its domain, since it is a combination of the functions sinxand cosx(both of which are continuous for allx) using the operation . The domain of tanxis all values ofxexcept those where cosx= 0, that is all values ofxexcept the odd multiples of2 Example: Removable DiscontinuityRecall that last day we found limx!0x2sin(1=x) using the squeeze theorem. What is the limit?Does the function
n(x) =x2sin(1=x)x >0 x2sin(1=x)x <0
have a removable discontinuity at zero? (in other words can I dene the function to have a value atx= 0 making a continuous function?) n1(x) =8
:x2sin(1=x)x >0
?x= 0 x2sin(1=x)x <0
Using continuity to calculate limits.NoteIf a functionf(x) is continuous on its domain and ifais in the domain off, then
lim x!af(x) =f(a): That is, ifais in the domain off, we can calculate the limit ataby evaluation. Ifais not in the domain off, we can sometimes use the methods discussed in the last lecture to determine if the limit exists or nd its value.4Example(a) Find the following limit:
lim x!2 xcos2xx+ sinxComposition of functions
Please reviewLecture 13 on Composing Functionson the Alegra/Precalculus review page. We can further expand our catalogue of functions continuous on their domains by considering com-position of functions.Theorem 3IfGis a continuous ataandFis continuous atG(a), then the composite function
FGgiven by (FG)(x) =F(G(x)) is continuous ata, and lim x!a(FG)(x) = (FG)(a):That is :
limx!aF(G(x)) =F(limx!aG(x)):Note that when the above conditions are met, we can calculate the limit by direct substitution.Recall that the domain ofFGis the set of pointsfx2domGjG(x)2domFg. Using this and the
theorem above we get:Iff(x) =F(G(x)), thenfis continuous at all points in its domain ifGis continuous at all points
in its domain andFis continuous at all points in its domain. ( Note that we can repeat the process to get the same result for a function of the formF(G(H(x))). )ExampleEvaluate the following limit: lim x!1rx2+x1x+ 3
LetG(x) =x2+x1x+3andF(x) =px. BothFandGare continuous on their domains, thereforeF(G(x)) =rx
2+x1x+ 3
is continuous on its domain. Sincex= 1 is in the domain, we can calculate the above limit by direct substitution. lim x!1rx2+x1x+ 3=r1 + 111 + 3
=r1 4 =12 5 Example(a) Find the domain of the following function and determine if it is continuous on its domain?: f(x) = cos(x3+ 1): Recall : IfG(x) =x3+ 1 andF(x) = cosx, thenF(G(x)) = cos(x3+ 1). The domain is thefx2Dom.GjG(x)2Dom.Fg.
(b) What is lim x!5cos(x3+ 1)?Intermediate Value TheoremHere is another interesting and useful property of functions which are continuous on a closed interval
[a;b]: the function must run through every y-value betweenf(a) andf(b). This makes sense since, acontinuous function can be drawn without lifting the pen from the paper.Intermediate value TheoremSuppose thatf(x) is continuous on the closed interval [a;b] and
letNbe any number betweenf(a) andf(b) (f(a)6=f(b)), then there exists a numbercin theinterval (a;b) withf(c) =N.ExampleIf we consider the functionf(x) =x21, on the interval [0;2], we see thatf(0) =1<0
andf(2) = 3>0. Therefore the intermediate value theorem says that there must be some number, c, between 0 and 2 withf(c) = 0. The graph off(x) crosses thexaxis at the point wherex=c. What is the value ofcin this case? f(x) =x21 = 0;ifx2= 1 This is true ifx=1, therefore forc= 12[0;2], we havef(c) = 0. In the above case, there is only one suchc, however in generalcmay not be unique. Also it may be dicult to determine the value ofc, however the theorem can be used to narrow down where the roots of an equation are. Exampleuse the intermediate value theorem to show that there is a root of the equation in the specied interval: cosx=x2(0;) 6 Extra Examples, Please attempt the following problems before looking at the solutions ExampleWhich of the following functions are continuous on the interval (0;1): f(x) =x3+x1x+ 2; g(x) =x2+ 3cosx; h(x) =px2+ 1x2; k(x) =jsinxj:
ExampleWhich of the following functions have a removable discontinuity atx= 2?: f(x) =x3+x1x2; g(x) =x24x2; h(x) =px2+ 1x2:
ExampleFind the domain of the following function and use Theorems 1, 2 and 3 to show that it is continuous on its domain: k(x) =3pcosxx10:ExampleEvaluate the following limits:
lim x!3p2 + cosxlim x!2 3 psinxx2 ExampleWhat is the domain of the following function and what are the (largest) intervals on which it is continuous? g(x) =1p1px Exampleuse the intermediate value theorem to show that there is a root of the equation in the specied interval:3px= 1x(0;1):
7Solutions
ExampleWhich of the following functions are continuous on the interval (0;1): f(x) =x3+x1x+ 2; g(x) =x2+ 3cosx; h(x) =px2+ 1x2; k(x) =jsinxj:
Sincef(x) is a rational function, it is continuous everywhere except atx=2, Therefore it is continuous
on the interval (0;1). By Theorem 2 and the continuity of polynomials and trigonometric functions,g(x) is continuous except where cosx= 0. Since cosx= 0 forx=2 ;32 ;:::, we haveg(x) is not continuous on (0;1). By theorems 2 and 3,h(x) is continuous everywhere except atx= 2. In factx= 2 is not in the domain of this function. Hence the function is not continuous on the interval (0;1). Sincek(x) =jsinxj=F(G(x)), whereG(x) = sinxandF(x) =jxj, we have thatk(x) is continuous everywhere on its domain since bothFandGare both continuous everywhere on their domains. Its not dicult to see that the domain ofkis all real numbers, hencekis continuous everywhere. (What does its graph look like?) ExampleWhich of the following functions have a removable discontinuity atx= 2?: f(x) =x3+x1x2; g(x) =x24x2; h(x) =px2+ 1x2:
lim x!2f(x) does not exist, since limx!2(x3+x1) = 9 and limx!2(x2) = 0. Therefore the discontinuity is not removable. lim x!2g(x) = limx!2(x2)(x+2)x2= limx!2(x+ 2) = 4. Therefore the discontinuity atx= 2 is removable by dening a piecewise function: g1(x) =g(x)x6= 2
4x= 2 lim x!2h(x) does not exist, since limx!2(px2+ 1) =p(5) and limx!2(x2) = 0. Therefore the
discontinuity is not removable. ExampleFind the domain of the following function and use Theorems 1, 2 and 3 to show that it is continuous on its domain: k(x) =3pcosxx10: The domain of this function is all values ofxexceptx= 10, since cosxis dened everywhere as is the cubed root function. Theorem 1 says that the cosine function is continuous everywhere and theorem 3 says thatf(x) =3pcosxis continuous for all real numbers since the cubed root function is continuous everywhere. Now we see from Theorem 2 thatk(x) =f(x)g(x)is continuous everywhere except where g(x) =x10 = 0, that is atx= 10.ExampleEvaluate the following limits:
lim x!3p2 + cosxlim x!2 3 psinxx2 SinceG(x) = 2 + cosxandF(x) =3pxare continuous everywhere, we haveF(Gx)) is continuous on its domain and we can calculate the rst limit by evaluation: lim x!3p2 + cosx=3p2 + cos=3p21 = 1: 8As above, we have
3psinxis continuous on its domain, therefore limx!2
3psinx=3psin
2 = 1. Since lim x!2 (x2 ) = 0, we have3psinxx2 approaches1in absolute value asxapproaches2 . Asx!2 sin(x)>0, hence3psinx >0. Asx!2 ,x2 <0, therefore the quotient has negative values and lim x!2 3 psinxx2 =1: ExampleWhat is the domain of the following function and what are the (largest) intervals on which it is continuous? g(x) =1p1px The domain of this function is allxwherep1px6= 0, i.e. allxwherex6= 1. By theorems 3 and 2,the function is continuous everywhere on its domain, therefore it is continuous on the intervals (1;1)
and (1;1). Exampleuse the intermediate value theorem to show that there is a root of the equation in the specied interval:3px= 1x(0;1):
Letg(x) =3px1 +x. We haveg(0) =1<0 andg(1) = 1>0. therefore by the intermediate value theorem, there is some numbercwith 0< c <1 for whichg(c) = 0. That is 3 pc= 1c as desired. 9 proofsContinuity of Trigonometric Functions
We see the following limits geometrically:
limx!0sin(x) = 0;limx!0cos(x) = 1:1.41.210.80.60.40.2-0.2-0.4-0.6-0.8-1-1.2-1.4-1-0.50.511cos(x)sin(x)xTheorem 2The functions sinxand cosxare continuous on the interval (1;1). In particular
; for any real numbera, we can evaluate the limits below by direct substitution lim x!asinx= sina;limx!acosx= cosa:ProofWe can use the addition formulas: sin(x+y) = sinxcosy+ cosxsiny;cos(x+y) = cosxcosysinxsiny; to show that lim x!asin(x) =sin(a) and limx!acos(x) =sin(a) for all real numbersa(radians). We will work through the details in the case of sin(x). lim x!asinx= limh!0sin(a+h) = limh!0(sinacosh+ cosasinh) = limh!0sinalimh!0cosh+ limh!0cosalimh!0sinh: = (sina)1 + (cosa)0 = sina:Proof of Theorem 2 (5) Theorem 2Iffandgare continuous ataandcis constant, then the following functions are also continuous ata: