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.

Theorem $\PageIndex{1}$ (sequential criterion of continuity).

(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).

Theorem $\PageIndex{2}$ (Cauchy criterion for functions).

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.

Theorem $\PageIndex{3}$

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 re-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).

Example $\PageIndex{1}$

(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

$$\lim _{z \rightarrow 0}(1+z)^{1 / z}=e.$$