Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

4.2: Some General Theorems on Limits and Continuity

( \newcommand{\kernel}{\mathrm{null}\,}\)

I. In §1 we gave the so-called "ε,δ" definition of continuity. Now we present another (equivalent) formulation, known as the sequential one. Roughly, it states that f is continuous iff it carries convergent sequences {xm}Df into convergent "image sequences" {f(xm)}. More precisely, we have the following theorem.

Theorem 4.2.1 (sequential criterion of continuity).

(i) A function

f:A(T,ρ), with A(S,ρ),

is continuous at a point pA iff for every sequence {xm}A such that xmp in (S,ρ), we have f(xm)f(p) in (T,ρ). In symbols,

({xm}A|xmp)f(xm)f(p).

(ii) Similarly, a point qT is a limit of f at p(pS) iff

({xm}A{p}|xmp)f(xm)q.

Note that in (2') we consider only sequences of terms other than p.

Proof

We first prove (ii). Suppose q is a limit of f at p, i.e. (see §1),

(ε>0)(δ>0)(xAG¬p(δ))f(x)Gq(ε).

Thus, given ε>0, there is δ>0 (henceforth fixed) such that

f(x)Gq(ε) whenever xA,xp, and xGp(δ).

We want to deduce (2'). Thus we fix any sequence

{xm}A{p},xmp.

Then

(m)xmA and xmp,

and Gp(δ) contains all but finitely many xm. Then these xm satisfy the conditions stated in (3). Hence f(xm)Gq(ε) for all but finitely many m. As ε is arbitrary, this implies f(xm)q (by the definition of limmf(xm)), as is required in (2'). Thus (2) (2').

Conversely, suppose (2) fails, i.e., its negation holds. (See the rules for forming negations of such formulas in Chapter 1, §§1-3.) Thus

(ε>0)(δ>0)(xAG¬p(δ))f(x)Gq(ε)

by the rules for quantifiers. We fix an ε satisfying (4), and let

δm=1m,m=1,2,

By (4), for each δm there is xm(depending on δm) such that

xmAG¬p(1m)

and

f(xm)Gq(ε),m=1,2,3,

We fix these xm. As xmA and xmp, we obtain a sequence

{xm}A{p}.

Also, as xmGp(1m), we have ρ(xm,p)<1/m0, and hence xmp. On the other hand, by (6), the image sequence {f(xm)} canverge to q (why?), i.e., (2') fails. Thus we see that (2') fails or holds accordingly as (2) does.

This proves assertion (ii). Now, by setting q=f(p) in (2) and (2'), we also obtain the first clause of the theorem, as to continuity.

Note 1. The theorem also applies to relative limits and continuity over a path B (just replace A by B in the proof), as well as to the cases p=± and q=± in E (for E can be treated as a metric space; see the end of Chapter 3, §11).

If the range space (T,ρ) is complete (Chapter 3, §17), then the image sequences {f(xm)} converge iff they are Cauchy. This leads to the following corollary.

Corollary 1. Let (T,ρ) be complete, such as En. Let a map f:AT with A(S,ρ) and a point pS be given. Then for f to have a limit at p, it suffices that {f(xm)} be Cauchy in (T,ρ) whenever {xm}A{p} and xmp in (S,ρ).

Indeed, as noted above, all such {f(xm)} converge. Thus it only remains to show that they tend to one and the same limit q, as is required in part (ii) of Theorem 1. We leave this as an exercise (Problem 1 below).

Theorem 4.2.2 (Cauchy criterion for functions).

With the assumptions of Corollary 1, the function f has a limit at p iff for each ε>0, there is δ>0 such that

ρ(f(x),f(x))<ε for all x,xAG¬p(δ).

In symbols,

(ε>0)(δ>0)(x,xAG¬p(δ))ρ(f(x),f(x))<ε.

Proof

Assume (7). To show that f has a limit at p, we use Corollary 1. Thus we take any sequence

{xm}A{p} with xmp

and show that {f(xm)} is Cauchy, i.e.,

(ε>0)(k)(m,n>k)ρ(f(xm),f(xn))<ε.

To do this, fix an arbitrary ε>0. By (7), we have

(x,xAG¬p(δ))ρ(f(x),f(x))<ε,

for some δ>0. Now as xmp, there is k such that

(m,n>k)xm,xnGp(δ).

As {xm}A{p}, we even have xm,xnAG¬p(δ). Hence by (7'),

(m,n>k)ρ(f(xm),f(xn))<ε;

i.e., {f(xm)} is Cauchy, as required in Corollary 1, and so f has a limit at p. This shows that (7) implies the existence of that limit.

The easy converse proof is left to the reader. (See Problem 2.)

II. Composite Functions. The composite of two functions

f:ST and g:TU,

denoted

gf(in that order),

is by definition a map of S into U given by

(gf)(x)=g(f(x)),xS.

Our next theorem states, roughly, that gf is continuous if g and f are. We shall use Theorem 1 to prove it.

Theorem 4.2.3

Let (S,ρ),(T,ρ), and (U,ρ) be metric spaces. If a function f:ST is continuous at a point pS, and if g:TU is continuous at the point q=f(p), then the composite function gf is continuous at p.

Proof

The domain of gf is S. So take any sequence

{xm}S with xmp.

As f is continuous at p, formula (1') yields f(xm)f(p), where f(xm) is in T=Dg. Hence, as g is continuous at f(p), we have

g(f(xm))g(f(p)), i.e., (gf)(xm)(gf)(p),

and this holds for any {xm}S with xmp. Thus gf satisfies condition (1') and is continuous at p.

Caution: The fact that

limxpf(x)=q and limyqg(y)=r

does not imply

limxpg(f(x))=r

(see Problem 3 for counterexamples).

Indeed, if {xm}S{p} and xmp, we obtain, as before, f(xm)q, but not f(xm)q. Thus we cannot re-apply formula (2') to obtain g(f(xm))r since (2') requires that f(xm)q. The argument still works if g is continuous at q (then (1') applies) or if f(x) never equals q then f(xm)q. It even suffices that f(x)q for x in some deleted globe about p( (see §1, Note 4). Hence we obtain the following corollary.

Corollary 2. With the notation of Theorem 3, suppose

f(x)q\)as\(xp, and g(y)r as yq.

Then

g(f(x))r as xp,

provided, however, that

(i) g is continuous at q, or

(ii) f(x)q for x in some deleted globe about p, or

(iii) f is one to one, at least when restricted to some G¬p(δ).

Indeed, (i) and (ii) suffice, as was explained above. Thus assume (iii). Then f can take the value q at most once, say, at some point

x0G¬p(δ).

As x0p, let

δ=ρ(x0,p)>0.

Then x0G¬p(δ), so f(x)q on G¬p(δ), and case (iii) reduces to (ii).

We now show how to apply Corollary 2.

Note 2. Suppose we know that

r=limyqg(y) exists.

Using this fact, we often pass to another variable x, setting y=f(x) where f is such that q=limxpf(x) for some p. We shall say that the substitution (or "change of variable") y=f(x) is admissible if one of the conditions (i), (ii), or (iii) of Corollary 2 holds. Then by Corollary 2,

limyqg(y)=r=limxpg(f(x))

(yielding the second limit).

Example 4.2.1

(A) Let

h(x)=(1+1x)x for |x|1.

Then

limx+h(x)=e.

For a proof, let n=f(x)=[x] be the integral part of x. Then for x>1,

(1+1n+1)nh(x)(1+1n)n+1.( Verify! )

As x+,n tends to + over integers, and by rules for sequences,

limn(1+1n)n+1=limn(1+1n)(1+1n)n=1limn(1+1n)n=1e=e,

with e as in Chapter 3, §15. Similarly one shows that also

limn(1+1n+1)n=e.

Thus (8) implies that also limx+h(x)=e (see Problem 6 below).

Remark. Here we used Corollary 2(ii) with

f(x)=[x],q=+, and g(n)=(1+1n)n.

The substitution n=f(x) is admissible since f(x)=n never equals +, its limit, thus satisfying Corollary 2(ii).

(B) Quite similarly, one shows that also

limx(1+1x)x=e.

See Problem 5.

(C) In Examples (A) and (B), we now substitute x=1/z. This is admissible by Corollary 2(ii) since the dependence between x and z is one to one. Then

z=1x0+ as x+, and z0 as x.

Thus (A) and (B) yield

limz0+(1+z)1/z=limz0(1+z)1/z=e.

Hence by Corollary 3 of §1, we obtain

limz0(1+z)1/z=e.


This page titled 4.2: Some General Theorems on Limits and Continuity is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?