CONTINUOUS FUNCTIONS
Section 5.2 Combinations of Continuous Functions
14. LetA:¼ ð0;1Þand letk:A!Rbe defined as follows. Forx2A;xirrational, we define kðxÞ ¼0; forx2Arational and of the formx¼m=nwith natural numbersm,nhaving no common factors except 1, we definekðxÞ:¼n. Prove thatkis unbounded on every open interval inA. Conclude thatkis not continuous at any point ofA. (See Example 5.1.6(h).)
15. Let f:ð0;1Þ !R be bounded but such that lim
x!0f does not exist. Show that there are two sequencesðxnÞandðynÞin (0, 1) with limðxnÞ ¼0¼limðynÞ, but such thatðfðxnÞÞandðfðynÞÞ exist but are not equal.
Remark To define quotients, it is sometimes more convenient to proceed as follows. If w:A!R, letA1:¼ fx2A:wðxÞ 6¼0g. We can define the quotientf=won the setA1by
ð1Þ f
w ðxÞ:¼ fðxÞ
wðxÞ for x2A1:
Ifw is continuous at a pointc2A1, it is clear that the restrictionw1 of w toA1is also continuous at c. Therefore it follows from Theorem 5.2.1(b) applied to w1 that f=w1 is continuous at c2A. Since ðf=wÞðxÞ ¼ ðf=w1ÞðxÞ for x2A1 it follows that f=w is continuous at c2A1. Similarly, iff andw are continuous on A, then the functionf=w, defined onA1by (1), is continuous onA1.
5.2.3 Examples (a) Polynomial functions.
Ifpis a polynomial function, so thatpðxÞ ¼anxnþan1xn1þ þa1xþa0for all x2R, then it follows from Example 4.2.5(f) that pðcÞ ¼lim
x!cp for any c2R. Thus a polynomial function is continuous onR.
(b) Rational functions.
Ifp and q are polynomial functions on R, then there are at most a finite number a1;. . .;amof real roots ofq. Ifx2 f= a1;. . .;amgthenqðxÞ 6¼0 so that we can define the rational functionrby
rðxÞ:¼pðxÞ
qðxÞ for x2 f= a1;. . .;amg: It was seen in Example 4.2.5(g) that if qðcÞ 6¼0, then
rðcÞ ¼pðcÞ qðcÞ¼lim
x!c
pðxÞ qðxÞ¼lim
x!crðxÞ:
In other words,ris continuous atc. Sincecis any real number that is not a root ofq, we infer thata rational function is continuous at every real number for which it is defined. (c) We shall show that the sine function sin is continuous onR.
To do so we make use of the following properties of the sine and cosine functions.
(See Section 8.4.) For allx,y,z2R we have:
jsin zj jzj; jcoszj 1; sinxsiny¼2 sin12ðxyÞ
cos12ðxþyÞ : Hence ifc2R, then we have
jsin xsincj 212jxcj 1¼ jxcj:
Therefore sin is continuous atc. Sincec2Ris arbitrary, it follows that sin is continuous onR.
(d) The cosine function is continuous on R.
We make use of the following properties of the sine and cosine functions. For all x;y;z2R we have:
jsinzj jzj; jsin zj 1; cosxcosy¼ 2 sin12ðxþyÞ
sin12ðxyÞ : Hence ifc2R, then we have
jcosxcoscj 2112jcxj ¼ jxcj:
5.2 COMBINATIONS OF CONTINUOUS FUNCTIONS 131
Therefore cos is continuous atc. Sincec2Ris arbitrary, it follows that cos is continuous onR. (Alternatively, we could use the relation cosx¼sinðxþp=2Þ.)
(e) The functions tan, cot, sec, csc are continuous where they are defined.
For example, the cotangent function is defined by cotx:¼cosx
sinx
provided sinx6¼0 (that is, providedx6¼np;n2Z). Since sin and cos are continuous on R, it follows (see the Remark before Example 5.2.3) that the function cot is continuous on its domain. The other trigonometric functions are treated similarly. &
5.2.4 Theorem Let AR,let f:A!R, and letjfjbe defined byjfjðxÞ:¼ jfðxÞjfor x2A.
(a) If f is continuous at a point c2A,thenjfjis continuous at c. (b) If f is continuous on A, thenjfjis continuous on A.
Proof. This is an immediate consequence of Exercise 4.2.14. Q.E.D.
5.2.5 Theorem Let AR,let f :A!R,and let fðxÞ 0for all x2A.We let ffiffiffi pf defined for x2A by ffiffiffi be
pf
ð ÞðxÞ:¼ ffiffiffiffiffiffiffiffiffi fðxÞ
p .
(a) If f is continuous at a point c2A, then ffiffiffi pf
is continuous at c. (b) If f is continuous on A, then ffiffiffi
pf
is continuous on A.
Proof. This is an immediate consequence of Exercise 4.2.15. Q.E.D.
Composition of Continuous Functions
We now show that if the functionf :A!Ris continuous at a pointcand ifg:B!Ris continuous atb¼f(c), then the compositiongf is continuous atc. In order to assure that gf is defined on all ofA, we also need to assume thatfðAÞ B.
5.2.6 Theorem Let A, BRand let f :A!Rand g:B!R be functions such that fðAÞ B. If f is continuous at a point c2A and g is continuous at b¼fðcÞ 2B, then the composition gf :A!R is continuous at c.
Proof. Let W be an e-neighborhood of g(b). Since g is continuous at b, there is a d-neighborhoodVofb¼fðcÞsuch that ify2B\VthengðyÞ 2W. Sincefis continuous at c, there is a g-neighborhood U of c such that if x2A\U, then fðxÞ 2V. (See Figure 5.2.1.) Since fðAÞ B, it follows that ifx2A\U, then fðxÞ 2B\V so that gfðxÞ ¼gðfðxÞÞ 2W. But sinceWis an arbitrarye-neighborhood ofg(b), this implies
thatgf is continuous atc. Q.E.D.
5.2.7 Theorem Let A;BR,let f :A!Rbe continuous on A, and let g:B!Rbe continuous on B. If fðAÞ B, then the composite function gf:A!Ris continuous on A. Proof. The theorem follows immediately from the preceding result, if f and g are continuous at every point ofAandB, respectively. Q.E.D.
Theorems 5.2.6 and 5.2.7 are very useful in establishing that certain functions are continuous. They can be used in many situations where it would be difficult to apply the definition of continuity directly.
5.2.8 Examples (a) Letg1ðxÞ:¼ jxjforx2R. It follows from the Triangle Inequality that
g1ðxÞ g1ðcÞ
j j jxcj
for all x;c2R. Hence g1 is continuous at c2R. If f :A!R is any function that is continuous onA, then Theorem 5.2.7 implies that g1f ¼ jfjis continuous onA. This gives another proof of Theorem 5.2.4.
(b) Let g2ðxÞ:¼ ffiffiffi px
for x0. It follows from Theorems 3.2.10 and 5.1.3 that g2is continuous at any numberc0. Iff :A!Ris continuous onAand iffðxÞ 0 for all x2A, then it follows from Theorem 5.2.7 thatg2f ¼ ffiffiffi
pf
is continuous onA. This gives another proof of Theorem 5.2.5.
(c) Letg3ðxÞ:¼sinxforx2R. We have seen in Example 5.2.3(c) thatg3is continuous onR. Iff :A!Ris continuous onA, then it follows from Theorem 5.2.7 thatg3f is continuous onA.
In particular, iffðxÞ:¼1=xforx6¼0, then the functiongðxÞ:¼sinð1=xÞis continu- ous at every pointc6¼0. [We have seen, in Example 5.1.8(a), thatgcannot be defined at
0 in order to become continuous at that point.] &
Exercises for Section 5.2
1. Determine the points of continuity of the following functions and state which theorems are used in each case.
(a) fðxÞ:¼x2þ2xþ1
x2þ1 ðx2RÞ; (b) gðxÞ:¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xþpffiffiffix
p ðx0Þ;
(c) hðxÞ:¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ jsinxj p
x ðx6¼0Þ; (d) kðxÞ:¼cos ffiffiffiffiffiffiffiffiffiffiffiffiffi 1þx2
p ðx2RÞ:
2. Show that iff :A!Ris continuous onARand ifn2N, then the functionfndefined by fnðxÞ ¼ ðfðxÞÞn, forx2A, is continuous onA.
3. Give an example of functionsfandgthat are both discontinuous at a pointcinRsuch that (a) the sumfþgis continuous atc, (b) the productfgis continuous atc.
Figure 5.2.1 The composition offandg
5.2 COMBINATIONS OF CONTINUOUS FUNCTIONS 133
4. Letx7!vxbdenote the greatest integer function (see Exercise 5.1.4). Determine the points of continuity of the functionfðxÞ:¼xvxb;x2R.
5. Letgbe defined onRbygð1Þ:¼0, andgðxÞ:¼2 ifx6¼1, and letfðxÞ:¼xþ1 for allx2R. Show that lim
x!0gf6¼ ðgfÞð0Þ. Why doesn’t this contradict Theorem 5.2.6?
6. Letf,gbe defined onRand letc2R. Suppose that lim
x!cf ¼band thatgis continuous atb. Show that lim
x!cgf¼gðbÞ. (Compare this result with Theorem 5.2.7 and the preceding exercise.)
7. Give an example of a functionf:½0;1 !Rthat is discontinuous at every point of [0, 1] but such thatjfjis continuous on [0, 1].
8. Letf,gbe continuous fromRtoR, and suppose thatfðrÞ ¼gðrÞfor all rational numbersr. Is it true thatfðxÞ ¼gðxÞfor allx2R?
9. Leth:R!Rbe continuous onR satisfyinghðm=2nÞ ¼0 for allm2Z;n2N. Show that hðxÞ ¼0 for allx2R.
10. Letf:R!Rbe continuous onR, and letP:¼ fx2R:fðxÞ>0g. Ifc2P, show that there exists a neighborhoodVdðcÞ P.
11. Iffandgare continuous onR, letS:¼ fx2R:fðxÞ gðxÞg. IfðsnÞ Sand limðsnÞ ¼s, show thats2S.
12. A functionf:R!Ris said to beadditiveiffðxþyÞ ¼fðxÞ þfðyÞfor allx,yinR. Prove that if f is continuous at some point x0, then it is continuous at every point of R. (See Exercise 4.2.12.)
13. Suppose thatfis a continuous additive function onR. Ifc:¼fð1Þ, show that we havefðxÞ ¼cx for allx2R. [Hint: First show that ifris a rational number, thenfðrÞ ¼cr.]
14. Letg:R!R satisfy the relationgðxþyÞ ¼gðxÞgðyÞfor allx,yinR. Show that ifgis continuous atx¼0, thengis continuous at every point ofR. Also if we havegðaÞ ¼0 for some a2R, thengðxÞ ¼0 for allx2R.
15. Letf;g:R!Rbe continuous at a pointc, and lethðxÞ:¼supffðxÞ;gðxÞgforx2R. Show that hðxÞ ¼12ðfðxÞ þgðxÞÞ þ12jfðxÞ gðxÞj for all x2R. Use this to show that h is continuous atc.