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.
(i) A function
f:A→(T,ρ′), with A⊆(S,ρ),
is continuous at a point p∈A iff for every sequence {xm}⊆A such that xm→p in (S,ρ), we have f(xm)→f(p) in (T,ρ′). In symbols,
(∀{xm}⊆A|xm→p)f(xm)→f(p).
(ii) Similarly, a point q∈T is a limit of f at p(p∈S) iff
(∀{xm}⊆A−{p}|xm→p)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)(∀x∈A∩G¬p(δ))f(x)∈Gq(ε).
Thus, given ε>0, there is δ>0 (henceforth fixed) such that
f(x)∈Gq(ε) whenever x∈A,x≠p, and x∈Gp(δ).
We want to deduce (2'). Thus we fix any sequence
{xm}⊆A−{p},xm→p.
Then
(∀m)xm∈A and xm≠p,
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 limm→∞f(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)(∃x∈A∩G¬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
xm∈A∩G¬p(1m)
and
f(xm)∉Gq(ε),m=1,2,3,…
We fix these xm. As xm∈A and xm≠p, we obtain a sequence
{xm}⊆A−{p}.
Also, as xm∈Gp(1m), we have ρ(xm,p)<1/m→0, and hence xm→p. 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:A→T with A⊆(S,ρ) and a point p∈S 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 xm→p 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).
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,x′∈A∩G¬p(δ).
In symbols,
(∀ε>0)(∃δ>0)(∀x,x′∈A∩G¬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 xm→p
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,x′∈A∩G¬p(δ))ρ′(f(x),f(x′))<ε,
for some δ>0. Now as xm→p, there is k such that
(∀m,n>k)xm,xn∈Gp(δ).
As {xm}⊆A−{p}, we even have xm,xn∈A∩G¬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:S→T and g:T→U,
denoted
g∘f(in that order),
is by definition a map of S into U given by
(g∘f)(x)=g(f(x)),x∈S.
Our next theorem states, roughly, that g∘f is continuous if g and f are. We shall use Theorem 1 to prove it.
Let (S,ρ),(T,ρ′), and (U,ρ′′) be metric spaces. If a function f:S→T is continuous at a point p∈S, and if g:T→U is continuous at the point q=f(p), then the composite function g∘f is continuous at p.
- Proof
-
The domain of g∘f is S. So take any sequence
{xm}⊆S with xm→p.
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., (g∘f)(xm)→(g∘f)(p),
and this holds for any {xm}⊆S with xm→p. Thus g∘f satisfies condition (1') and is continuous at p. ◻
Caution: The fact that
limx→pf(x)=q and limy→qg(y)=r
does not imply
limx→pg(f(x))=r
(see Problem 3 for counterexamples).
Indeed, if {xm}⊆S−{p} and xm→p, 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\(x→p, and g(y)→r as y→q.
Then
g(f(x))→r as x→p,
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
x0∈G¬p(δ).
As x0≠p, let
δ′=ρ(x0,p)>0.
Then x0∉G¬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=limy→qg(y) exists.
Using this fact, we often pass to another variable x, setting y=f(x) where f is such that q=limx→pf(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,
limy→qg(y)=r=limx→pg(f(x))
(yielding the second limit).
(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)n≤h(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=1⋅limn→∞(1+1n)n=1⋅e=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=1x→0+ as x→+∞, and z→0− as x→−∞.
Thus (A) and (B) yield
limz→0+(1+z)1/z=limz→0−(1+z)1/z=e.
Hence by Corollary 3 of §1, we obtain
limz→0(1+z)1/z=e.