
# 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

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