LIMITS
Section 4.2 Limit Theorems
We shall now obtain results that are useful in calculating limits of functions. These results are parallel to the limit theorems established in Section 3.2 for sequences. In fact, in most cases these results can be proved by using Theorem 4.1.8 and results from Section 3.2.
Alternatively, the results in this section can be proved by usinge-darguments that are very similar to the ones employed in Section 3.2.
4.2.1 Definition LetAR, letf :A!R, and letc2Rbe a cluster point ofA. We say thatfisbounded on a neighborhood ofcif there exists ad-neighborhoodVdð Þc ofcand a constantM>0 such that we havejf xð Þj M for allx2A\Vdð Þc .
4.2.2 Theorem If ARand f :A!Rhas a limit at c2R,then f is bounded on some neighborhood of c.
Proof. IfL:¼lim
x!cf, then fore¼1, there existsd>0 such that if 0<jxcj<d, then f xð Þ L
j j<1; hence (by Corollary 2.2.4(a)), f xð Þ
j j j j L jf xð Þ Lj<1:
Therefore, ifx2A\Vdð Þ;c x6¼c, thenjf xð Þj j j þL 1. Ifc2= A, we takeM ¼j j þL 1, while ifc2Awe takeM:¼supfjf cð Þj;j j þL 1g. It follows that if x2A\Vdð Þc, then
f xð Þ
j j M. This shows thatfis bounded on the neighborhood Vdð Þc of c. Q.E.D.
The next definition is similar to the definition for sums, differences, products, and quotients of sequences given in Section 3.2.
4.2.3 Definition LetARand letfandgbe functions defined onAtoR. We define the sumf þg, thedifferencef g, and theproductfgonAtoRto be the functions given by
fþg
ð Þð Þx :¼f xð Þ þg xð Þ; ðfgÞð Þx :¼f xð Þ g xð Þ;
f g
ð Þð Þx :¼f xð Þg xð Þ
for allx2A. Further, if b2R, we define themultiple bfto be the function given by bf
ð Þð Þx :¼bf xð Þ for all x2A:
Finally, ifh xð Þ 6¼0 forx2A, we define thequotientf=hto be the function given by f
h
x
ð Þ:¼f xð Þ
h xð Þ for all x2A:
4.2.4 Theorem Let AR,let f and g be functions on A toR,and let c2Rbe a cluster point of A. Further, let b2R.
(a) If lim
x!cf ¼L andlim
x!cg¼M,then:
limx!cðfþgÞ ¼LþM; lim
x!cðf gÞ ¼LM;
limx!cð Þ ¼f g LM; lim
x!cð Þ ¼bf bL:
(b) If h:A!R,if h xð Þ 6¼0for all x2A,and iflim
x!ch¼H6¼0, then
x!clim f h
¼ L H:
4.2 LIMIT THEOREMS 111
Proof. One proof of this theorem is exactly similar to that of Theorem 3.2.3. Alterna- tively, it can be proved by making use of Theorems 3.2.3 and 4.1.8. For example, let (xn) be any sequence inAsuch thatxn6¼cforn2N, andc¼limð Þxn . It follows from Theorem 4.1.8 that
limðf xð Þn Þ ¼L; limðg xð Þn Þ ¼M:
On the other hand, Definition 4.2.3 implies that f g
ð Þð Þ ¼xn f xð Þg xn ð Þn for n2N:
Therefore an application of Theorem 3.2.3 yields limðð Þf g ð Þxn Þ ¼ limðf xð Þg xn ð Þn Þ
¼½limðf xð Þn Þ½limðg xð Þn Þ ¼LM: Consequently, it follows from Theorem 4.1.8 that
x!climð Þ ¼f g limðð Þf gð Þxn Þ ¼LM:
The other parts of this theorem are proved in a similar manner. We leave the details to
the reader. Q.E.D.
Remark LetAR, and letf1;f2;. . .; fnbe functions onAtoR, and letcbe a cluster point of A. If Lk:¼lim
x!cfk for k¼1;. . .;n;then it follows from Theorem 4.2.4 by an Induction argument that
L1þL2þ þLn ¼lim
x!cðf1þf2þ þfnÞ;
and
L1L2 Ln¼limðf1f2 fnÞ:
In particular, we deduce that ifL¼lim
x!cfandn2N, then Ln¼lim
x!cðf xð ÞÞn:
4.2.5 Examples (a) Some of the limits that were established in Section 4.1 can be proved by using Theorem 4.2.4. For example, it follows from this result that since
x!climx¼c, then lim
x!cx2¼c2, and that ifc>0, then limx!c
1 x¼ 1
x!climx¼1 c: (b) lim
x!2x2þ1 x34
¼20.
It follows from Theorem 4.2.4 that
x!lim2x2þ1 x34
¼ lim
x!2x2þ1
x!lim2x34
¼54¼20: (c) lim
x!2
x34 x2þ1
¼4 5.
If we apply Theorem 4.2.4(b), we have limx!2
x34 x2þ1¼lim
x!2x34
x!lim2x2þ1¼4 5: Note that since the limit in the denominator [i.e., lim
x!2x2þ1
¼5] is not equal to 0, then Theorem 4.2.4(b) is applicable.
(d) lim
x!2¼x24 3x6¼4
3.
If we letf xð Þ:¼x24 andh xð Þ:¼3x6 forx2R, then wecannotuse Theorem 4.2.4(b) to evaluate lim
x!2ðf xð Þ=h xð ÞÞbecause H¼lim
x!2h xð Þ ¼lim
x!2ð3x6Þ ¼326¼0: However, ifx6¼2, then it follows that
x24
3x6¼ðxþ2Þðx2Þ 3ðx2Þ ¼1
3ðxþ2Þ:
Therefore we have
x!lim2
x24 3x6¼lim
x!2
1
3ðxþ2Þ ¼1 3 lim
x!2xþ2
¼4 3:
Note that the functiong xð Þ ¼ðx24Þ=ð3x6Þhas a limit atx¼2even though it is not defined there.
(e) lim
x!0
1
xdoes not exist in R. Of course lim
x!01¼1 and H:¼lim
x!0x¼0. However, since H ¼ 0, wecannot use Theorem 4.2.4(b) to evaluate lim
x!0ð1=xÞ. In fact, as was seen in Example 4.1.10(a), the functionwð Þ ¼x 1=x does not have a limit atx ¼0. This conclusion also follows from Theorem 4.2.2 since the functionwð Þ ¼x 1=xis not bounded on a neighborhood ofx¼0.
(f) Ifpis a polynomial function, then lim
x!cp xð Þ ¼p cð Þ.
Letpbe a polynomial function onRso thatp xð Þ ¼anxnþan1xn1þ þa1xþ a0 for allx2R. It follows from Theorem 4.2.4 and the fact that lim
x!cxk¼ck that limx!cp xð Þ ¼ lim
x!c anxnþan1xn1þ þa1xþa0
¼ lim
x!cðanxnÞ þlim
x!can1xn1
þ þlim
x!cða1xÞ þlim
x!ca0
¼ ancnþan1cn1þ þa1cþa0
¼ p cð Þ:
Hence lim
x!cp xð Þ ¼p cð Þ for any polynomial functionp.
(g) Ifpandqare polynomial functions onR and ifq cð Þ 6¼0, then limx!c
p xð Þ q xð Þ¼p cð Þ
q cð Þ:
Sinceq(x) is a polynomial function, it follows from a theorem in algebra that there are at most a finite number of real numbersa1;. . .;am[the real zeroes ofq(x)] such thatq aj ¼0 and such that ifx2= fa1;. . .;amg, thenq xð Þ 6¼0. Hence, ifx2= fa1;. . .;amg, we can define
r xð Þ:¼p xð Þ q xð Þ:
4.2 LIMIT THEOREMS 113
Ifcis not a zero ofq(x), thenq cð Þ 6¼0, and it follows from part (f) that lim
x!cq xð Þ ¼q cð Þ 6¼0.
Therefore we can apply Theorem 4.2.4(b) to conclude that limx!c
p xð Þ q xð Þ¼lim
x!cp xð Þ limx!cq xð Þ¼p cð Þ
q cð Þ: &
The next result is a direct analogue of Theorem 3.2.6.
4.2.6 Theorem Let AR,let f :A!R,and let c2R be a cluster point of A. If af xð Þ b f or all x2A;x6¼c;
and iflim
x!cf exists, then alim
x!cf b. Proof. Indeed, if L¼lim
x!cf, then it follows from Theorem 4.1.8 that if (xn) is any sequence of real numbers such that c6¼xn2A for all n2N and if the sequence (xn) converges to c, then the sequence ðf xð Þn Þ converges to L. Sinceaf xð Þ n b for all n2N, it follows from Theorem 3.2.6 thataLb. Q.E.D.
We now state an analogue of the Squeeze Theorem 3.2.7. We leave its proof to the reader.
4.2.7 Squeeze Theorem Let AR,let f, g, h: A!R,and let c2Rbe a cluster point of A. If
f xð Þ g xð Þ h xð Þ f or all x2A;x6¼c;
and iflim
x!cf ¼L¼lim
x!ch,then lim
x!cg¼L. 4.2.8 Examples (a) lim
x!0x3=2¼0ðx>0Þ.
Letf xð Þ:¼x3=2 forx>0. Since the inequalityx<x1=21 holds for 0<x1 (why?), it follows thatx2f xð Þ ¼x3=2xfor 0<x1. Since
x!lim0x2¼0 and lim
x!0x¼0; it follows from the Squeeze Theorem 4.2.7 that lim
x!0x3=2¼0.
(b) lim
x!0sinx¼0.
It will be proved later (see Theorem 8.4.8), that xsinxx for all x0: Since lim
x!0ðxÞ ¼0, it follows from the Squeeze Theorem that lim
x!0sinx¼0.
(c) lim
x!0cosx¼1.
It will be proved later (see Theorem 8.4.8) that
ð1Þ 112x2cosx1 for all x2R:
Since lim
x!0112x2
¼1, it follows from the Squeeze Theorem that lim
x!0cosx¼1.
(d) lim
x!0
cosx1 x
¼0.
We cannot use Theorem 4.2.4(b) to evaluate this limit. (Why not?) However, it follows from the inequality (1) in part (c) that
12xðcosx1Þ=x0 for x>0 and that
0ðcosx1Þ=x 12x for x<0:
Now letf xð Þ:¼ x=2 forx0 andf xð Þ:¼0 forx<0, and leth xð Þ:¼0 forx0 and
h xð Þ:¼ x=2 forx<0. Then we have
f xð Þ ðcosx1Þ=xh xð Þ for x6¼0: Since it is readily seen that lim
x!0f ¼0¼lim
x!0h, it follows from the Squeeze Theorem that
x!lim0ðcosx1Þ=x¼0.
(e) lim
x!0
sinx x
¼1.
Again we cannot use Theorem 4.2.4(b) to evaluate this limit. However, it will be proved later (see Theorem 8.4.8) that
x16x3sinxx for x0 and that
xsinxx16x3 for x0: Therefore it follows (why?) that
116x2ðsinxÞ=x1 for all x6¼0: But since lim
x!0116x2
¼116lim
x!0x2 ¼1, we infer from the Squeeze Theorem that
x!lim0ðsinxÞ=x¼1.
(f) lim
x!0ðxsin 1ð =xÞÞ ¼0.
Letf xð Þ ¼xsin 1ð =xÞ forx6¼0. Since 1sinz1 for all z2R, we have the inequality
j j x f xð Þ ¼xsin 1ð =xÞ j jx for allx2R,x6¼0. Since lim
x!0j j ¼x 0, it follows from the Squeeze Theorem that lim
x!0f ¼0.
For a graph, see Figure 5.1.3 or the cover of this book. &
There are results that are parallel to Theorems 3.2.9 and 3.2.10; however, we will leave them as exercises. We conclude this section with a result that is, in some sense, a partial converse to Theorem 4.2.6.
4.2.9 Theorem Let AR,let f :A!R and let c2R be a cluster point of A. If
x!climf >0 respectively; lim
x!cf <0
h i
;
then there exists a neighborhood Vd(c)of c such that f(x)>0 [respectively, f(x)<0]for all x2A\Vdð Þc ,x6¼c.
Proof. LetL:¼lim
x!cf and suppose thatL>0. We takee¼12L>0 in Definition 4.1.4, and obtain a number d >0 such that if 0<jxcj<d and x2A, then jf xð Þ Lj<12L. Therefore (why?) it follows that if x2A\Vdð Þc,x6¼c, thenf xð Þ>12L>0.
IfL<0, a similar argument applies. Q.E.D.
4.2 LIMIT THEOREMS 115
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 .