LIMITS
Section 4.1 Limits of Functions
4.1 LIMITS OF FUNCTIONS 105
0<jxcj<1, we have jf xð Þ bj ¼jbbj ¼0<e. Since e>0 is arbitrary, we conclude from Definition 4.1.4 that lim
x!cf xð Þ ¼b. (b) lim
x!cx¼c.
Letg xð Þ:¼xfor allx2R. Ife>0, we choosedð Þe :¼e. Then if 0<jxcj<dð Þe, we have jg xð Þ cj ¼jxcj<e. Sincee>0 is arbitrary, we deduce that lim
x!cg¼c. (c) lim
x!cx2¼c2:
Leth xð Þ:¼x2 for allx2R. We want to make the difference h xð Þ c2
¼x2c2
less than a preassignede>0 by takingx sufficiently close toc. To do so, we note that x2c2¼ðxþcÞðxcÞ. Moreover, ifjxcj<1, then
x
j j<j j þc 1 so that jxþcj j j þx j jc <2j j þc 1: Therefore, ifjxcj<1, we have
ð1Þ x2c2¼jxþcjjxcj<ð2j j þc 1Þjxcj:
Moreover this last term will be less than e provided we take jxcj<e=ð2j j þc 1Þ. Consequently, if we choose
dð Þe :¼inf 1; e 2j j þc 1
;
then if 0<jxcj<dð Þe, it will follow first that jxcj<1 so that (1) is valid, and therefore, sincejxcj<e=ð2j j þc 1Þthat
x2c2
<ð2j j þc 1Þjxcj<e:
Since we have a way of choosingdð Þe >0 for an arbitrary choice ofe>0, we infer that
x!climh xð Þ ¼lim
x!cx2¼c2. (d) lim
x!c
1 x¼1
c ifc>0.
Letwð Þx :¼1=xforx>0 and letc>0. To show that lim
x!cw¼1=cwe wish to make the difference
wð Þ x 1 c
¼ 1 x1
c
less than a preassignede>0 by takingxsufficiently close toc>0. We first note that 1
x1 c
¼ 1
cxðcxÞ
¼ 1 cxjxcj
for x > 0. It is useful to get an upper bound for the term 1=(cx) that holds in some neighborhood ofc. In particular, if jxcj<12c, then12c<x<32c(why?), so that
0< 1 cx<2
c2 for jxcj<1 2c:
Therefore, for these values of xwe have
ð2Þ wð Þ x 1
c
2 c2jxcj:
In order to make this last term less thaneit suffices to takejxcj<12c2e. Consequently, if we choose
dð Þe :¼inf 1 2c; 1
2c2e
;
then if 0<jxcj<dð Þe, it will follow first thatjxcj<12cso that (2) is valid, and therefore, sincejxcj<12c2
e, that wð Þ x 1
c
¼ 1 x1
c
<e:
Since we have a way of choosingdð Þe >0 for an arbitrary choice ofe>0, we infer that limx!cw¼1=c.
(e) lim
x!2
x34 x2þ1¼4
5:
Let cð Þx :¼ðx34Þ=ðx2þ1Þ for x2R. Then a little algebraic manipulation gives us
cð Þ x 4 5
¼5x34x224 5ðx2þ1Þ
¼5x3þ6xþ12
5ðx2þ1Þ jx2j:
To get a bound on the coefficient ofjx2j, we restrictxby the condition 1<x<3.
For x in this interval, we have 5x2þ6xþ12532þ63þ12¼75 and 5ðx2þ1Þ 5 1ð þ1Þ ¼10, so that
cð Þ x 4 5
75
10jx2j ¼15 2 jx2j:
Now for givene>0, we choose
dð Þe :¼inf 1; 2 15e
:
Then if 0<jx2j<dð Þe, we havejcð Þ x ð4=5Þj ð15=2Þjx2j<e. Sincee>0 is
arbitrary, the assertion is proved. &
Sequential Criterion for Limits
The following important formulation of limit of a function is in terms of limits of sequences. This characterization permits the theory of Chapter 3 to be applied to the study of limits of functions.
4.1.8 Theorem (Sequential Criterion) Let f :A!R and let c be a cluster point of A.
Then the following are equivalent. (i) lim
x!cf ¼L:
(ii) For every sequence(xn)in A that converges to c such that xn6¼c for all n2N,the sequence (f(xn))converges to L.
Proof. (i))(ii). Assume fhas limitL atc, and suppose (xn) is a sequence inAwith limð Þ ¼xn candxn6¼cfor alln. We must prove that the sequenceðf xð Þn Þconverges toL. Lete>0 be given. Then by Definition 4.1.4, there existsd>0 such that ifx2Asatisfies 4.1 LIMITS OF FUNCTIONS 107
0<jxcj<d, then f(x) satisfies jf xð Þ Lj<e. We now apply the definition of convergent sequence for the given d to obtain a natural number K(d) such that if n>
Kð Þd thenjxncj<d. But for each suchxnwe havejf xð Þ n Lj<e. Thus ifn>Kð Þd , thenjf xð Þ n Lj<e. Therefore, the sequenceðf xð Þn Þconverges toL.
(ii))(i). [The proof is a contrapositive argument.] If (i) is not true, then there exists ane0-neighborhoodVe0ð ÞL such that no matter whatd-neighborhood ofcwe pick, there will be at least one numberxdinA\Vdð Þc withxd6¼csuch thatf xð Þd 2= Ve0ð ÞL . Hence for everyn2N, the (1=n)-neighborhood ofccontains a numberxnsuch that
0<jxncj<1=n and xn2A; but such that
f xð Þ n L
j j e0 for all n2N:
We conclude that the sequence (xn) inAn{c} converges toc, but the sequenceðf xð Þn Þdoes not converge toL. Therefore we have shown that if (i) is not true, then (ii) is not true. We
conclude that (ii) implies (i). Q.E.D.
We shall see in the next section that many of the basic limit properties of functions can be established by using corresponding properties for convergent sequences. For example, we know from our work with sequences that if (xn) is any sequence that converges to a number c, then x2n converges to c2. Therefore, by the sequential criterion, we can conclude that the functionh xð Þ:¼x2 has limit lim
x!ch xð Þ ¼c2. Divergence Criteria
It is often important to be able to show (i) that a certain number isnotthe limit of a function at a point, or (ii) that the functiondoes not havea limit at a point. The following result is a consequence of (the proof of) Theorem 4.1.8. We leave the details of its proof as an important exercise.
4.1.9 Divergence Criteria Let AR, let f :A!R and let c2R be a cluster point of A.
(a) If L2R,then f doesnothave limit L at c if and only if there exists a sequence(xn)in A with xn6¼c for all n2N such that the sequence (xn)converges to c but the sequence
f xð Þn
ð Þdoes notconverge to L.
(b) The function f doesnothave a limit at c if and only if there exists a sequence(xn)in A with xn6¼c for all n2N such that the sequence (xn)converges to c but the sequence
f xð Þn
ð Þdoes notconverge inR.
We now give some applications of this result to show how it can be used.
4.1.10 Examples (a) lim
x!0ð1=xÞdoes not exist in R.
As in Example 4.1.7(d), letwð Þx :¼1=xforx>0. However, here we considerc¼0.
The argument given in Example 4.1.7(d) breaks down if c¼0 since we cannot obtain a bound such as that in (2) of that example. Indeed, if we take the sequence (xn) with xn :¼1=n for n2N, then limð Þ ¼xn 0, but wð Þ ¼xn 1=ð1=nÞ ¼n. As we know, the sequence ðwð Þxn Þ ¼ð Þn is not convergent in R, since it is not bounded. Hence, by Theorem 4.1.9(b), lim
x!0ð1=xÞdoes not exist inR. (b) lim
x!0sgnð Þx does not exist.
Let thesignum functionsgn be defined by
sgnð Þx :¼ þ1 for x>0; 0 for x¼0; 1 for x<0: 8<
:
Note that sgnð Þ ¼x x=j jx forx6¼0. (See Figure 4.1.2.) We shall show that sgn does not have a limit atx¼0. We shall do this by showing that there is a sequence (xn) such that limð Þ ¼xn 0, but such that sgnð ð Þxn Þdoes not converge.
Indeed, letxn:¼ ð Þ1 n=n forn2N so that limð Þ ¼xn 0. However, since sgnð Þ ¼ xn ð Þ1 n for n2N;
it follows from Example 3.4.6(a) that sgnð ð Þxn Þdoes not converge. Therefore lim
x!0sgnð Þx does not exist.
(c)y lim
x!0sin 1ð =xÞdoes not exist inR.
Letg xð Þ:¼sin 1ð =xÞforx6¼0. (See Figure 4.1.3.) We shall show thatgdoes not have a limit atc¼0, by exhibiting two sequences (xn) and (yn) withxn6¼0 andyn6¼0 for all n2Nand such that limð Þ ¼xn 0 and limð Þ ¼yn 0, but such that limðg xð Þn Þ 6¼limðg yð Þn Þ. In view of Theorem 4.1.9 this implies that lim
x!0g cannot exist. (Explain why.)
Indeed, we recall from calculus that sint¼0 ift¼npforn2Z, and that sint¼ þ1 ift¼12pþ2pnforn2Z. Now letxn:¼1=npforn2N; then limð Þ ¼xn 0 andg xð Þ ¼n
sinnp¼0 for all n2N, so that limðg xð Þn Þ ¼0. On the other hand, let yn :¼
1
2pþ2pn
1
for n2N; then limð Þ ¼yn 0 and g yð Þ ¼n sin12pþ2pn
¼1 for all n2N, so that limðg yð Þn Þ ¼1. We conclude that lim
x!0sin 1ð =xÞdoes not exist. &
yIn order to have some interesting applications in this and later examples, we shall make use of well-known properties of trigonometric and exponential functions that will be established in Chapter 8.
Figure 4.1.3 The functiong xð Þ ¼sin 1ð =xÞðx6¼0Þ Figure 4.1.2 The signum function