LIMITS
Section 4.3 Some Extensions of the Limit Concept y
Exercises for Section 4.2
1. Apply Theorem 4.2.4 to determine the following limits:
(a) lim
x!1ðxþ1Þð2xþ3Þ ðx2RÞ; (b) lim
x!1
x2þ2
x22 ðx>0Þ;
(c) lim
x!2
1 xþ1 1
2x
x>0
ð Þ; (d) lim
x!0
xþ1
x2þ2 ðx2RÞ:
2. Determine the following limits and state which theorems are used in each case. (You may wish to use Exercise 15 below.)
(a) lim
x!2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2xþ1 xþ3 r
x>0
ð Þ; (b) lim
x!2
x24
x2 ðx>0Þ;
(c) lim
x!0
xþ1 ð Þ21
x ðx>0Þ; (d) lim
x!1
ffiffiffix p 1
x1 ðx>0Þ:
3. Find lim
x!0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ2x
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ3x p
xþ2x2 wherex>0.
4. Prove that lim
x!0cos 1=xð Þdoes not exist but that lim
x!0xcos 1=xð Þ ¼0.
5. Letf,gbe defined onARtoR, and letcbe a cluster point ofA. Suppose thatfis bounded on a neighborhood ofcand that lim
x!cg¼0. Prove that lim
x!cf g¼0.
6. Use the definition of the limit to prove the first assertion in Theorem 4.2.4(a).
7. Use the sequential formulation of the limit to prove Theorem 4.2.4(b).
8. Letn2Nbe such thatn3. Derive the inequalityx2xnx2for1<x<1. Then use the fact that lim
x!0x2¼0 to show that lim
x!0xn¼0.
9. Letf,gbe defined onAtoRand letcbe a cluster point ofA.
(a) Show that if both lim
x!cf and lim
x!cðfþgÞexist, then lim
x!cgexists.
(b) If lim
x!cf and lim
x!cf gexist, does it follow that lim
x!cgexists?
10. Give examples of functionsfandgsuch thatfandgdo not have limits at a pointc, but such that bothfþgandfghave limits atc.
11. Determine whether the following limits exist inR. (a) lim
x!0sin 1=x 2
x6¼0
ð Þ, (b) lim
x!0xsin 1=x 2
x6¼0 ð Þ, (c) lim
x!0sgn sin 1ð =xÞ ðx6¼0Þ, (d) lim
x!0
ffiffiffix
p sin 1 =x2
x>0 ð Þ. 12. Letf:R!Rbe such thatf xð þyÞ ¼f xð Þ þf yð Þfor allx,yinR. Assume that lim
x!0f ¼L exists. Prove thatL¼0, and then prove thatfhas a limit at every pointc2R. [Hint: First note that fð Þ ¼2x f xð Þ þf xð Þ ¼2f xð Þforx2R. Also note thatf xð Þ ¼f xð cÞ þf cð Þforx,cinR.]
13. Functionsfandgare defined onRbyf(x) :¼xþ1 andg(x) :¼2 ifx6¼1 andg(1) :¼0.
(a) Find lim
x!1g(f(x)) and compare with the value of g(lim
x!1f(x)).
(b) Find lim
x!1f(g(x)) and compare with the value off(lim
x!1g(x)).
14. LetAR, letf:A!Rand letc2Rbe a cluster point ofA. If lim
x!cfexists, and ifj jf denotes the function defined forx2Abyj jf ð Þx :¼jf xð Þj, prove that lim
x!cj j ¼f lim
x!cf .
15. Let AR, letf :A!R, and let c2R be a cluster point ofA. In addition, suppose that f xð Þ 0 for allx2A, and let ffiffiffi
pf
be the function defined forx2Aby ffiffiffi pf
ð Þ(x) :¼ ffiffiffiffiffiffiffiffiffi f xð Þ p . If
x!climf exists, prove that lim
x!c
ffiffiffif
p ¼ ffiffiffiffiffiffiffiffiffiffi limx!cf
q .
4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 117
One-Sided Limits
There are times when a function f may not possess a limit at a point c, yet a limit does exist when the function is restricted to an interval on one side of the cluster pointc.
For example, the signum function considered in Example 4.1.10(b), and illustra- ted in Figure 4.1.2, has no limit atc¼0. However, if we restrict the signum function to the interval (0, 1), the resulting function has a limit of 1 at c¼ 0. Similarly, if we restrict the signum function to the interval (1, 0), the resulting function has a limit of1 at c¼0. These are elementary examples of right-hand and left-hand limits at c¼0.
4.3.1 Definition LetA2R and letf :A!R.
(i) Ifc2Ris a cluster point of the setA\ðc; 1Þ ¼fx2A:x>cg, then we say that L2R is aright-hand limit off atcand we write
x!cþlim f ¼L or lim
x!cþf xð Þ ¼L
if given any e>0 there exists a d¼dð Þe >0 such that for all x2A with 0<xc<d, then jf xð Þ Lj<e.
(ii) Ifc2Ris a cluster point of the setA\ 1;ð cÞ ¼fx2A:x<cg, then we say that L2R is aleft-hand limit offatcand we write
x!clim f ¼L or lim
x!cf xð Þ ¼L
if given anye>0 there exists ad>0 such that for allx2Awith 0<cx<d, thenjf xð Þ Lj<e.
Notes (1) The limits lim
x!cþfand lim
x!cf are calledone-sided limits offatc. It is possible that neither one-sided limit may exist. Also, one of them may exist without the other existing. Similarly, as is the case forf xð Þ:¼sgnð Þx atc¼0, they may both exist and be different.
(2) IfAis an interval with left endpointc, then it is readily seen thatf :A!Rhas a limit atcif and only if it has a right-hand limit atc. Moreover, in this case the limit lim
x!cf and the right-hand limit lim
x!cþf are equal. (A similar situation occurs for the left-hand limit whenA is an interval with right endpoint c.)
The reader can show thatfcan have only one right-hand (respectively, left-hand) limit at a point. There are results analogous to those established in Sections 4.1 and 4.2 for two- sided limits. In particular, the existence of one-sided limits can be reduced to sequential considerations.
4.3.2 Theorem Let AR,let f :A!R,and let c2Rbe a cluster point of A\ðc;1Þ. Then the following statements are equivalent:
(i) lim
x!cþf ¼L.
(ii) For every sequence(xn)that converges to c such that xn2A and xn>c for all n2N,the sequence f xð ð Þn Þconverges to L.
We leave the proof of this result (and the formulation and proof of the analogous result for left-hand limits) to the reader. We will not take the space to write out the formulations of the one-sided version of the other results in Sections 4.1 and 4.2.
The following result relates the notion of the limit of a function to one-sided limits. We leave its proof as an exercise.
4.3.3 Theorem Let AR,let f :A!R,and let c2R be a cluster point of both of the sets A\ðc;1Þ and A\ 1;ð cÞ. Then lim
x!cf ¼L if and only if
x!cþlim f ¼L¼ lim
x!cf.
4.3.4 Examples (a) Letf xð Þ:¼sgnð Þx .
We have seen in Example 4.1.10(b) that sgn does not have a limit at 0. It is clear that
x!lim0þsgnð Þ ¼ þx 1 and that lim
x!0sgnð Þ ¼ x 1. Since these one-sided limits are different, it also follows from Theorem 4.3.3 that sgn(x) does not have a limit at 0.
(b) Let g xð Þ:¼e1=xfor x6¼0. (See Figure 4.3.1.)
We first show thatg does not have a finite right-hand limit at c¼0 since it is not bounded on any right-hand neighborhood 0ð ; dÞ of 0. We shall make use of the inequality
ð1Þ 0<t<et for t>0;
which will be proved later (see Corollary 8.3.3). It follows from (1) that ifx>0, then 0<1=x<e1=x. Hence, if we take xn¼1=n, then g xð Þn >n for all n2N. Therefore
x!lim0þe1=x does not exist inR. However, lim
x!0e1=x¼0. Indeed, ifx <0 and we taket¼ 1=x in (1) we obtain 0<1=x<e1=x. Sincex<0, this implies that 0<e1=x<xfor allx<0. It follows from this inequality that lim
x!0e1=x¼0.
(c) Leth xð Þ:¼1=e1=xþ1
forx6¼0. (See Figure 4.3.2.)
We have seen in part (b) that 0<1=x<e1=xfor x>0, whence 0< 1
e1=xþ1< 1 e1=x<x;
which implies that lim
x!0þh¼0.
Figure 4.3.1 Graph of g xð Þ ¼e1=x ðx6¼0Þ
Figure 4.3.2 Graph of h xð Þ ¼1=e1=xþ1
x6¼0
ð Þ
Since we have seen in part (b) that lim
x!0e1=x¼0, it follows from the analogue of Theorem 4.2.4(b) for left-hand limits that
x!lim0
1 e1=xþ1
¼ 1
x!lim0e1=xþ1¼ 1 0þ1¼1:
Note that for this function, both one-sided limits exist inR, but they are unequal. &
Infinite Limits
The functionf xð Þ:¼1=x2forx6¼0 (see Figure 4.3.3) is not bounded on a neighborhood of 0, so it cannot have a limit in the sense of Definition 4.1.4. While the symbols 1 ¼ þ1ð Þand1do not represent real numbers, it is sometimes useful to be able to say that ‘‘f xð Þ ¼1=x2 tends to 1 as x!0.’’ This use of 1 will not cause any difficulties, provided we exercise caution andneverinterpret 1or1 as being real numbers.
4.3.5 Definition LetAR, letf:A!R, and let c2R be a cluster point of A. (i) We say thatftends to 1asx!c, and write
limx!cf ¼ 1;
if for every a2R there exists d¼d að Þ>0 such that for all x2A with 0<jxcj<d, thenf xð Þ>a.
(ii) We say thatftends to 1as x!c, and write limx!cf ¼ 1;
if for every b2R there exists d¼d bð Þ>0 such that for all x2A with 0<jxcj<d, thenf xð Þ<b.
4.3.6 Examples (a) lim
x!01=x2
¼ 1. For, ifa>0 is given, letd:¼1= ffiffiffi pa
. It follows that if 0<j jx <d, thenx2<1=aso that 1=x2>a.
(b) Let g xð Þ:¼1=xforx6¼0. (See Figure 4.3.4.) Figure 4.3.3 Graph of
f xð Þ ¼1=x2 ðx6¼0Þ Figure 4.3.4 Graph of g xð Þ ¼1=x ðx6¼0Þ
4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 119
The functiongdoesnottend to either1or1asx!0. For, ifa>0 thengðxÞ<a for allx<0, so thatgdoes not tend to1asx!0. Similarly, ifb<0 thengðxÞ>bfor all
x>0, so thatgdoes not tend to 1as x!0. &
While many of the results in Sections 4.1 and 4.2 have extensions to this limiting notion, not all of them do since 1 are not real numbers. The following result is an analogue of the Squeeze Theorem 4.2.7. (See also Theorem 3.6.4.)
4.3.7 Theorem Let AR, let f;g:A!R, and let c2R be a cluster point of A.
Suppose that fðxÞ gðxÞfor all x2A;x6¼c. (a) If lim
x!cf ¼ 1, then lim
x!cg¼ 1. (b) If lim
x!cg¼ 1, thenlim
x!cf ¼ 1. Proof. (a) If lim
x!cf ¼ 1 and a2R is given, then there exists dðaÞ>0 such that if 0<jxcj <dðaÞandx2A, thenfðxÞ>a. But sincefðxÞ gðxÞfor allx2A;x6¼c, it follows that if 0<jxcj<dðaÞandx2A, thengðxÞ>a. Therefore lim
x!cg¼ 1.
The proof of (b) is similar. Q.E.D.
The functiongðxÞ ¼1=xconsidered in Example 4.3.6(b) suggests that it might be useful to consider one-sided infinite limits. We will define only right-hand infinite limits.
4.3.8 Definition Let AR and let f :A!R. If c2R is a cluster point of the set A\ ðc;1Þ ¼ fx2A:x>cg, then we say that f tends to 1 [respectively, 1] as x!cþ, and we write
x!cþlim f ¼ 1h
respectively; lim
x!cþf ¼ 1i
;
if for everya2Rthere isd¼dðaÞ>0 such that for allx2Awith 0<xc<d, then fðxÞ>a[respectively,fðxÞ<a].
4.3.9 Examples (a) LetgðxÞ:¼1=xforx6¼0. We have noted in Example 4.3.6(b) that
x!lim0g does not exist. However, it is an easy exercise to show that
x!lim0þð1=xÞ ¼ 1 and lim
x!0ð1=xÞ ¼ 1:
(b) It was seen in Example 4.3.4(b) that the function gðxÞ:¼e1=x for x6¼0 is not bounded on any interval 0ð ;dÞ;d>0. Hence the right-hand limit ofe1=xasx!0þdoes not exist in the sense of Definition 4.3.1(i). However, since
1=x<e1=x for x>0; it is readily seen that lim
x!0þe1=x¼ 1in the sense of Definition 4.3.8. &
Limits at Infinity
It is also desirable to define the notion of the limit of a function asx! 1. The definition as x! 1is similar.
4.3.10 Definition Let AR and let f :A!R. Suppose that ða;1Þ A for some a2R. We say thatL2R is alimit off asx! 1, and write
x!1limf ¼L or lim
x!1fðxÞ ¼L;
if given anye>0 there existsK¼KðeÞ>asuch that for anyx>K, thenjfðxÞ Lj<e. The reader should note the close resemblance between 4.3.10 and the definition of a limit of a sequence.
We leave it to the reader to show that the limits offasx! 1are unique whenever they exist. We also have sequential criteria for these limits; we shall only state the criterion as x! 1. This uses the notion of the limit of a properly divergent sequence (see Definition 3.6.1).
4.3.11 Theorem Let AR, let f :A!R, and suppose that ða;1Þ A for some a2R.Then the following statements are equivalent:
(i) L¼ lim
x!1f.
(ii) For every sequence(xn)in A\ ða;1Þsuch thatlimðxnÞ ¼ 1,the sequenceðfðxnÞÞ converges to L.
We leave it to the reader to prove this theorem and to formulate and prove the companion result concerning the limit asx! 1.
4.3.12 Examples (a) LetgðxÞ:¼1=xfor x6¼0.
It is an elementary exercise to show that lim
x!1ð1=xÞ ¼0¼ lim
x!1ð1=xÞ. (See Figure 4.3.4.)
(b) Let fðxÞ:¼1=x2 forx6¼0.
The reader may show that lim
x!1ð1=x2Þ ¼0¼ lim
x!1ð1=x2Þ. (See Figure 4.3.3.) One way to do this is to show that ifx1 then 01=x21=x. In view of part (a), this implies that lim
x!1ð1=x2Þ ¼0. &
Just as it is convenient to be able to say thatfðxÞ ! 1asx!cfor c2R, it is convenient to have the corresponding notion as x! 1. We will treat the case where x! 1.
4.3.13 Definition Let AR and let f :A!R. Suppose that ða;1Þ A for some a2A. We say thatftends to 1[respectively,1]asx! 1, and write
x!1lim f ¼ 1 respectively; lim
x!1f ¼ 1
h i
if given anya2R there exists K¼KðaÞ>a such that for anyx >K, thenfðxÞ>a [respectively,fðxÞ<a].
As before there is a sequential criterion for this limit.
4.3.14 Theorem Let A2R, let f :A!R, and suppose that ða;1Þ A for some a2R.Then the following statements are equivalent:
4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 121
(i) lim
x!1f ¼ 1 ½respectively; lim
x!1f ¼ 1.
(ii) For every sequence (xn)inða;1Þsuch thatlimðxnÞ ¼ 1,then limðf xð Þn Þ ¼ 1 respectively;limðf xð Þn Þ ¼ 1
½ .
The next result is an analogue of Theorem 3.6.5.
4.3.15 Theorem Let AR, let f;g:A!R,and suppose that ða;1Þ A for some a2R. Suppose further that gðxÞ>0for all x>a and that for some L2R;L6¼0, we have
x!1lim fðxÞ gðxÞ¼L:
(i) If L>0,then lim
x!1f ¼ 1if and only if lim
x!1g¼ 1. (ii) If L<0,then lim
x!1f ¼ 1if and only if lim
x!1g¼ 1.
Proof. (i) SinceL >0, the hypothesis implies that there existsa1>a such that 0<1
2LfðxÞ gðxÞ<3
2L for x>a1: Therefore we have1
2L
gðxÞ<fðxÞ<3
2L
gðxÞfor allx>a1, from which the conclusion follows readily.
The proof of (ii) is similar. Q.E.D.
We leave it to the reader to formulate the analogous result as x! 1. 4.3.16 Examples (a) lim
x!1xn¼ 1forn2N.
LetgðxÞ:¼xnforx2 ð0;1Þ. Givena2R, letK:¼supf1;ag. Then for allx>K, we have gðxÞ ¼xnx>a. Sincea2R is arbitrary, it follows that lim
x!1g¼ 1. (b) lim
x!1xn¼ 1forn2N,n even, and lim
x!1xn¼ 1 forn2N,nodd.
We will treat the casenodd, sayn¼2kþ1 withk¼0, 1, . . . . Givena2R, let K:¼inffa;1g. For anyx <K, then since ðx2Þk1, we havexn ¼ ðx2Þkxx<a.
Since a2R is arbitrary, it follows that lim
x!1xn¼ 1. (c) Letp:R!R be the polynomial function
pðxÞ:¼anxnþan1xn1þ þa1xþa0: Then lim
x!1p¼ 1ifan>0, and lim
x!1p¼ 1ifan<0.
Indeed, letgðxÞ:¼xn and apply Theorem 4.3.15. Since pðxÞ
gðxÞ¼anþan1
1 x
þ þa1
1 xn1
þa0
1 xn
; it follows that lim
x!1ðpðxÞ=gðxÞÞ ¼an. Since lim
x!1g¼ 1, the assertion follows from Theorem 4.3.15.
(d) Letpbe the polynomial function in part (c). Then lim
x!1p¼ 1[respectively,1] if n is even [respectively, odd] andan>0.
We leave the details to the reader. &
Exercises for Section 4.3
1. Prove Theorem 4.3.2.
2. Give an example of a function that has a right-hand limit but not a left-hand limit at a point.
3. LetfðxÞ:¼ jxj1=2forx6¼0. Show that lim
x!0þfðxÞ ¼ lim
x!0fðxÞ ¼ þ1.
4. Let c2R and letf be defined for x2 ðc;1Þand fðxÞ>0 for all x2 ðc;1Þ. Show that
x!climf ¼ 1if and only if lim
x!c1=f¼0.
5. Evaluate the following limits, or show that they do not exist.
(a) lim
x!1þ
x
x1 ðx6¼1Þ, (b) lim
x!1
x
x1 ðx6¼1Þ, (c) lim
x!0þðxþ2Þ= ffiffiffi px
x>0
ð Þ, (d) lim
x!1ðxþ2Þ= ffiffiffi px
x>0 ð Þ, (e) lim
x!0
ffiffiffiffiffiffiffiffiffiffiffi xþ1 p
=x ðx>1Þ, (f) lim
x!1
ffiffiffiffiffiffiffiffiffiffiffi xþ1 p
=x ðx>0Þ, (g) lim
x!1
ffiffiffix p 5
ffiffiffix
p þ3 ðx>0Þ, (h) lim
x!1
ffiffiffix pffiffiffix px
þx ðx>0Þ. 6. Prove Theorem 4.3.11.
7. Suppose thatfandghave limits inRasx! 1and thatfðxÞ gðxÞfor allx2 ða;1Þ. Prove that lim
x!1f lim
x!1g.
8. Letfbe defined on (0,1) toR. Prove that lim
x!1fðxÞ ¼Lif and only if lim
x!0þfð1=xÞ ¼L.
9. Show that iff :ða;1Þ !Ris such that lim
x!1xfðxÞ ¼LwhereL2R, then lim
x!1fðxÞ ¼0.
10. Prove Theorem 4.3.14.
11. Suppose that lim
x!cfðxÞ ¼LwhereL>0, and that lim
x!cgðxÞ ¼ 1. Show that lim
x!cfðxÞgðxÞ ¼ 1.
IfL¼0, show by example that this conclusion may fail.
12. Find functions f and g defined on (0, 1) such that lim
x!1 f ¼ 1 and lim
x!1g¼ 1, and
x!1limðfgÞ ¼0. Can you find such functions, withg(x) >0 for allx2 ð0;1Þ, such that
x!1limf=g¼0?
13. Let f and g be defined on (a, 1) and suppose lim
x!1f¼L and lim
x!1g¼ 1. Prove that
x!1limf g¼L.
4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 123