4.2: Some General Theorems on Limits and Continuity
I. In §1 we gave the so-called "\(\varepsilon, \delta\)" 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 \(\left\{x_{m}\right\} \subseteq D_{f}\) into convergent "image sequences" \(\left\{f\left(x_{m}\right)\right\} .\) More precisely, we have the following theorem.
(i) A function
\[f : A \rightarrow\left(T, \rho^{\prime}\right),\text{ with } A \subseteq(S, \rho),\]
is continuous at a point \(p \in A\) iff for every sequence \(\left\{x_{m}\right\} \subseteq A\) such that \(x_{m} \rightarrow p\) in \((S, \rho),\) we have \(f\left(x_{m}\right) \rightarrow f(p)\) in \(\left(T, \rho^{\prime}\right) .\) In symbols,
\[\left(\forall\left\{x_{m}\right\} \subseteq A | x_{m} \rightarrow p\right) \quad f\left(x_{m}\right) \rightarrow f(p).\]
(ii) Similarly, a point \(q \in T\) is a limit of \(f\) at \(p(p \in S)\) iff
\[\left(\forall\left\{x_{m}\right\} \subseteq A-\{p\} | x_{m} \rightarrow p\right) \quad f\left(x_{m}\right) \rightarrow q.\]
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),
\[(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in A \cap G_{\neg p}(\delta)\right) \quad f(x) \in G_{q}(\varepsilon).\]
Thus, given \(\varepsilon>0,\) there is \(\delta>0\) (henceforth fixed) such that
\[f(x) \in G_{q}(\varepsilon)\text{ whenever } x \in A, x \neq p,\text{ and } x \in G_{p}(\delta).\]
We want to deduce (2'). Thus we fix any sequence
\[\left\{x_{m}\right\} \subseteq A-\{p\}, x_{m} \rightarrow p.\]
Then
\[(\forall m) \quad x_{m} \in A\text{ and } x_{m} \neq p,\]
and \(G_{p}(\delta)\) contains all but finitely many \(x_{m} .\) Then these \(x_{m}\) satisfy the conditions stated in (3) . Hence \(f\left(x_{m}\right) \in G_{q}(\varepsilon)\) for all but finitely many \(m .\) As \(\varepsilon\) is arbitrary, this implies \(f\left(x_{m}\right) \rightarrow q\) (by the definition of \(\lim _{m \rightarrow \infty} f\left(x_{m}\right) ),\) as is required in (2') . Thus (2) \(\Longrightarrow\) (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
\[(\exists \varepsilon>0)(\forall \delta>0)\left(\exists x \in A \cap G_{\neg p}(\delta)\right) \quad f(x) \notin G_{q}(\varepsilon)\]
by the rules for quantifiers. We fix an \(\varepsilon\) satisfying (4), and let
\[\delta_{m}=\frac{1}{m}, \quad m=1,2, \ldots\]
By (4) , for each \(\delta_{m}\) there is \(x_{m}\left(\text {depending on } \delta_{m}\right)\) such that
\[x_{m} \in A \cap G_{\neg p}\left(\frac{1}{m}\right)\]
and
\[f\left(x_{m}\right) \notin G_{q}(\varepsilon), \quad m=1,2,3, \ldots\]
We fix these \(x_{m} .\) As \(x_{m} \in A\) and \(x_{m} \neq p,\) we obtain a sequence
\[\left\{x_{m}\right\} \subseteq A-\{p\}.\]
Also, as \(x_{m} \in G_{p}\left(\frac{1}{m}\right),\) we have \(\rho\left(x_{m}, p\right)<1 / m \rightarrow 0,\) and hence \(x_{m} \rightarrow p\). On the other hand, by (6) , the image sequence \(\left\{f\left(x_{m}\right)\right\}\) 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. \(\square\)
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 \(p=\pm \infty\) and \(q=\pm \infty\) in \(E^{*}\) (for \(E^{*}\) can be treated as a metric space; see the end of Chapter 3, §11).
If the range space \(\left(T, \rho^{\prime}\right)\) is complete (Chapter 3, §17), then the image sequences \(\left\{f\left(x_{m}\right)\right\}\) converge iff they are Cauchy. This leads to the following corollary.
Corollary 1. Let \(\left(T, \rho^{\prime}\right)\) be complete, such as \(E^{n} .\) Let a map \(f : A \rightarrow T\) with \(A \subseteq(S, \rho)\) and a point \(p \in S\) be given. Then for \(f\) to have a limit at \(p,\) it suffices that \(\left\{f\left(x_{m}\right)\right\}\) be Cauchy in \(\left(T, \rho^{\prime}\right)\) whenever \(\left\{x_{m}\right\} \subseteq A-\{p\}\) and \(x_{m} \rightarrow p\) in \((S, \rho).\)
Indeed, as noted above, all such \(\left\{f\left(x_{m}\right)\right\}\) 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 \(\varepsilon>0,\) there is \(\delta>0\) such that
\[\rho^{\prime}\left(f(x), f\left(x^{\prime}\right)\right)<\varepsilon\text{ for all } x, x^{\prime} \in A \cap G_{\neg p}(\delta).\]
In symbols,
\[(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x, x^{\prime} \in A \cap G_{\neg p}(\delta)\right) \quad \rho^{\prime}\left(f(x), f\left(x^{\prime}\right)\right)<\varepsilon.\]
- Proof
-
Assume (7). To show that \(f\) has a limit at \(p,\) we use Corollary 1. Thus we take any sequence
\[\left\{x_{m}\right\} \subseteq A-\{p\}\text{ with } x_{m} \rightarrow p\]
and show that \(\left\{f\left(x_{m}\right)\right\}\) is Cauchy, i.e.,
\[(\forall \varepsilon>0)(\exists k)(\forall m, n>k) \quad \rho^{\prime}\left(f\left(x_{m}\right), f\left(x_{n}\right)\right)<\varepsilon.\]
To do this, fix an arbitrary \(\varepsilon>0.\) By (7), we have
\[\left(\forall x, x^{\prime} \in A \cap G_{\neg p}(\delta)\right) \quad \rho^{\prime}\left(f(x), f\left(x^{\prime}\right)\right)<\varepsilon,\]
for some \(\delta>0 .\) Now as \(x_{m} \rightarrow p,\) there is \(k\) such that
\[(\forall m, n>k) \quad x_{m}, x_{n} \in G_{p}(\delta).\]
As \(\left\{x_{m}\right\} \subseteq A-\{p\},\) we even have \(x_{m}, x_{n} \in A \cap G_{\neg p}(\delta).\) Hence by (7'),
\[(\forall m, n>k) \quad \rho^{\prime}\left(f\left(x_{m}\right), f\left(x_{n}\right)\right)<\varepsilon;\]
i.e., \(\left\{f\left(x_{m}\right)\right\}\) 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.) \(\square\)
II. Composite Functions. The composite of two functions
\[f : S \rightarrow T\text{ and } g : T \rightarrow U,\]
denoted
\[g \circ f \quad(\text {in that order}),\]
is by definition a map of \(S\) into \(U\) given by
\[(g \circ f)(x)=g(f(x)), \quad x \in S.\]
Our next theorem states, roughly, that \(g \circ f\) is continuous if \(g\) and \(f\) are. We shall use Theorem 1 to prove it.
Let \((S, \rho),\left(T, \rho^{\prime}\right),\) and \(\left(U, \rho^{\prime \prime}\right)\) be metric spaces. If a function \(f : S \rightarrow T\) is continuous at a point \(p \in S,\) and if \(g : T \rightarrow U\) is continuous at the point \(q=f(p),\) then the composite function \(g \circ f\) is continuous at \(p\).
- Proof
-
The domain of \(g \circ f\) is \(S .\) So take any sequence
\[\left\{x_{m}\right\} \subseteq S\text{ with } x_{m} \rightarrow p.\]
As \(f\) is continuous at \(p,\) formula (1') yields \(f\left(x_{m}\right) \rightarrow f(p),\) where \(f\left(x_{m}\right)\) is in \(T=D_{g} .\) Hence, as \(g\) is continuous at \(f(p),\) we have
\[g\left(f\left(x_{m}\right)\right) \rightarrow g(f(p)),\text{ i.e., } (g \circ f)\left(x_{m}\right) \rightarrow(g \circ f)(p),\]
and this holds for any \(\left\{x_{m}\right\} \subseteq S\) with \(x_{m} \rightarrow p.\) Thus \(g \circ f\) satisfies condition (1') and is continuous at \(p.\) \(\square\)
Caution: The fact that
\[\lim _{x \rightarrow p} f(x)=q\text{ and } \lim _{y \rightarrow q} g(y)=r\]
does not imply
\[\lim _{x \rightarrow p} g(f(x))=r\]
(see Problem 3 for counterexamples).
Indeed, if \(\left\{x_{m}\right\} \subseteq S-\{p\}\) and \(x_{m} \rightarrow p,\) we obtain, as before, \(f\left(x_{m}\right) \rightarrow q,\) but not \(f\left(x_{m}\right) \neq q .\) Thus we cannot r e-apply formula (2') to obtain \(g\left(f\left(x_{m}\right)\right) \rightarrow r\) since (2') requires that \(f\left(x_{m}\right) \neq q .\) The argument still works if \(g\) is continuous at \(q\) ( then (1') applies ) or if \(f(x)\) never equals \(q\) then \(f(x_{m}) \neq q\). It even suffices that \(f(x) \neq q\) 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) \rightarrow q\) as \(x \rightarrow p,\text{ and } g(y) \rightarrow r\text{ as } y \rightarrow q.\]
Then
\[g(f(x)) \rightarrow r\text{ as } x \rightarrow p,\]
provided, however, that
(i) \(g\) is continuous at \(q,\) or
(ii) \(f(x) \neq q\) for \(x\) in some deleted globe about \(p,\) or
(iii) \(f\) is one to one, at least when restricted to some \(G_{\neg p}(\delta)\).
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
\[x_{0} \in G_{\neg p}(\delta).\]
As \(x_{0} \neq p,\) let
\[\delta^{\prime}=\rho\left(x_{0}, p\right)>0.\]
Then \(x_{0} \notin G_{\neg p}\left(\delta^{\prime}\right),\) so \(f(x) \neq q\) on \(G_{\neg p}\left(\delta^{\prime}\right),\) and case (iii) reduces to (ii).
We now show how to apply Corollary 2.
Note 2. Suppose we know that
\[r=\lim _{y \rightarrow q} g(y)\text{ exists.}\]
Using this fact, we often pass to another variable \(x,\) setting \(y=f(x)\) where \(f\) is such that \(q=\lim _{x \rightarrow p} f(x)\) 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,
\[\lim _{y \rightarrow q} g(y)=r=\lim _{x \rightarrow p} g(f(x))\]
(yielding the second limit).
(A) Let
\[h(x)=\left(1+\frac{1}{x}\right)^{x}\text{ for } |x| \geq 1.\]
Then
\[\lim _{x \rightarrow+\infty} h(x)=e.\]
For a proof, let \(n=f(x)=[x]\) be the integral part of \(x.\) Then for \(x>1\),
\[\left(1+\frac{1}{n+1}\right)^{n} \leq h(x) \leq\left(1+\frac{1}{n}\right)^{n+1} . \quad(\text { Verify! })\]
As \(x \rightarrow+\infty, n\) tends to \(+\infty\) over integers, and by rules for sequences,
\[\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)^{n}=1 \cdot \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=1 \cdot e=e,\]
with \(e\) as in Chapter 3, §15. Similarly one shows that also
\[\lim _{n \rightarrow \infty}\left(1+\frac{1}{n+1}\right)^{n}=e.\]
Thus (8) implies that also \(\lim _{x \rightarrow+\infty} h(x)=e\) (see Problem 6 below).
Remark. Here we used Corollary 2(ii) with
\[f(x)=[x], q=+\infty,\text{ and } g(n)=\left(1+\frac{1}{n}\right)^{n}.\]
The substitution \(n=f(x)\) is admissible since \(f(x)=n\) never equals \(+\infty,\) its limit, thus satisfying Corollary 2(ii).
(B) Quite similarly, one shows that also
\[\lim _{x \rightarrow-\infty}\left(1+\frac{1}{x}\right)^{x}=e.\]
See Problem 5.
(C) In Examples \((\mathrm{A})\) and \((\mathrm{B}),\) 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
\[z=\frac{1}{x} \rightarrow 0^{+}\text{ as } x \rightarrow+\infty,\text{ and } z \rightarrow 0^{-}\text{ as } x \rightarrow-\infty.\]
Thus \((\mathrm{A})\) and \((\mathrm{B})\) yield
\[\lim _{z \rightarrow 0^{+}}(1+z)^{1 / z}=\lim _{z \rightarrow 0^{-}}(1+z)^{1 / z}=e.\]
Hence by Corollary 3 of §1, we obtain
\[\lim _{z \rightarrow 0}(1+z)^{1 / z}=e.\]