A sequence converging to a limit and a function's value "not jumping around" are really the same idea once the sequential definition of a functional limit is available. This chapter makes that idea precise, develops the standard toolkit (Lipschitz continuity, the sequential criterion, the algebra of continuous functions), and puts it to work: the intermediate value theorem, the extreme value theorem on compact sets, and a full proof of the fundamental theorem of algebra all follow from nothing more than the definition of continuity plus what sequences already do.
Definition of Continuity
Let \(D\subseteq\mathbb{R}\) (or \(D\subseteq\mathbb{C}\)) and \(f:D\to\mathbb{R}\). \(f\) is continuous at a point \(a\in D\) if
\[ \forall\epsilon>0\ \exists\delta>0\ \forall x\in D:\quad |x-a|<\delta \implies |f(x)-f(a)|<\epsilon. \]
\(f\) is continuous (on \(D\)) if it is continuous at every point of \(D\). Geometrically, no matter how narrow a target band of height \(2\epsilon\) around \(f(a)\) is drawn, some interval of width \(2\delta\) around \(a\) maps entirely inside it — the graph has no jump, hole, or vertical asymptote at \(a\).
Lipschitz Functions
\(f:D\to\mathbb{R}\) is Lipschitz (with constant \(L\)) if
\[ |f(x)-f(y)|\leq L|x-y|\quad\text{for all }x,y\in D. \]
Every Lipschitz function is continuous: given \(\epsilon>0\), choosing \(\delta:=\epsilon/L\) (or any \(\delta>0\) at all if \(L=0\)) immediately gives \(|f(x)-f(a)|\leq L|x-a|<L\delta=\epsilon\) whenever \(|x-a|<\delta\). The point of Lipschitz continuity is that the same \(\delta\) works everywhere at once, which is a strictly stronger requirement than continuity alone.
\(f(x)=|x|\) is Lipschitz with \(L=1\) on all of \(\mathbb{R}\), by the reverse triangle inequality \(\bigl||x|-|y|\bigr|\leq|x-y|\). A convergent power series is Lipschitz on any closed sub-disk strictly inside its radius of convergence, since its derivative series converges there too and is therefore bounded. Not every continuous function is Lipschitz, though: \(f(x)=\sqrt{x}\) is continuous on \([0,1]\), but
\[ \frac{f(x)-f(0)}{x-0}=\frac{1}{\sqrt{x}}\to\infty\quad\text{as }x\downarrow0, \]
so no single \(L\) bounds \(|\sqrt{x}-\sqrt{y}|/|x-y|\) near \(0\). On any sub-interval \([\alpha,\beta]\) with \(0<\alpha<\beta\), however, \(\sqrt{\cdot}\) is Lipschitz: multiplying by the conjugate gives \(|\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{\sqrt{x}+\sqrt{y}}\leq\frac{|x-y|}{2\sqrt{\alpha}}\), so \(L=\frac{1}{2\sqrt{\alpha}}\) works there.
The Sequential Criterion for Continuity
\(f\) is continuous at \(a\in D\) if and only if \(f(x_n)\to f(a)\) for every sequence \((x_n)\) in \(D\) with \(x_n\to a\).
Proof. Suppose \(f\) is continuous at \(a\) and \(x_n\to a\). Given \(\epsilon>0\), pick \(\delta>0\) from continuity; since \(x_n\to a\), eventually \(|x_n-a|<\delta\), hence eventually \(|f(x_n)-f(a)|<\epsilon\). So \(f(x_n)\to f(a)\). Conversely, suppose \(f\) is not continuous at \(a\): some \(\epsilon>0\) has no working \(\delta\), so for every \(n\) there is an \(x_n\in D\) with \(|x_n-a|<1/n\) yet \(|f(x_n)-f(a)|\geq\epsilon\). Then \(x_n\to a\) but \(f(x_n)\not\to f(a)\), the contrapositive of the claim.
Because this reduces continuity to ordinary sequence convergence, every rule from the algebra of limits transfers immediately: if \(f\) and \(g\) are continuous at \(a\), so are \(f+g\), \(f-g\), \(cf\), \(fg\), and (where \(g(a)\neq0\)) \(f/g\); and if \(f\) is continuous at \(a\) and \(g\) is continuous at \(f(a)\), then \(g\circ f\) is continuous at \(a\) (apply the sequential criterion twice: \(x_n\to a\Rightarrow f(x_n)\to f(a)\Rightarrow g(f(x_n))\to g(f(a))\)). In particular, every polynomial is continuous on \(\mathbb{R}\), every rational function is continuous wherever its denominator is nonzero, and \(\Re(f)\), \(\Im(f)\), \(|f|\) are continuous whenever \(f\) is.
The Intermediate Value Theorem
Let \(f:[a,b]\to\mathbb{R}\) be continuous with \(f(a)<y<f(b)\) (or \(f(b)<y<f(a)\)). Then \(f(c)=y\) for some \(c\in(a,b)\).
Proof. Assume \(f(a)<y<f(b)\) (the other case follows by applying this to \(-f\)). Let
\[ c:=\sup\{x\in[a,b]:f(x)\leq y\}. \]
The set is nonempty (\(a\) is in it) and bounded above by \(b\), so \(c\) exists and \(c\in[a,b]\). By definition of supremum, there is a sequence \(x_n\to c\) with \(f(x_n)\leq y\), so continuity gives \(f(c)=\lim f(x_n)\leq y\); in particular \(c\neq b\) since \(f(b)>y\). For \(x\in(c,b]\), \(x\) is not in the set, so \(f(x)>y\); letting \(x\downarrow c\) through such points and using continuity again gives \(f(c)\geq y\). Hence \(f(c)=y\), and \(c\neq a\) since \(f(a)<y\), so \(c\in(a,b)\).
Fixed Points, Roots, and Real n-th Roots
Fixed point theorem. A continuous \(f:[a,b]\to[a,b]\) has a fixed point: \(f(c)=c\) for some \(c\in[a,b]\). Indeed \(g(x):=f(x)-x\) is continuous with \(g(a)=f(a)-a\geq0\) and \(g(b)=f(b)-b\leq0\) (since \(f\) maps into \([a,b]\)); if either is \(0\) take \(c=a\) or \(c=b\), otherwise the intermediate value theorem gives \(g(c)=0\) for some \(c\in(a,b)\).
Existence of real \(n\)-th roots. For \(\alpha>0\) and \(n\in\mathbb{N}\), there is a (unique) \(\beta>0\) with \(\beta^n=\alpha\). \(f(x):=x^n\) is continuous, \(f(0)=0<\alpha\), and by Bernoulli's inequality \(f(1+\alpha)=(1+\alpha)^n\geq1+n\alpha>\alpha\), so the intermediate value theorem applied on \([0,1+\alpha]\) gives a \(\beta\) with \(\beta^n=\alpha\); uniqueness follows since \(f\) is strictly increasing on \([0,\infty)\).
Odd-degree real polynomials have a real root. Let \(p(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_0\) with \(n\) odd. For \(|x|\geq1\), \(|c_{n-1}x^{n-1}+\cdots+c_0|\leq(|c_{n-1}|+\cdots+|c_0|)|x|^{n-1}=:M|x|^{n-1}\), so \(p(x)=x^n(1+O(1/x))\) has the same sign as \(x^n\) once \(|x|>M+1\); since \(n\) is odd this makes \(p(R)>0\) and \(p(-R)<0\) for \(R:=M+1\), and the intermediate value theorem gives a root in \((-R,R)\).
Strictly Monotone Inverse Functions
A continuous injective \(f:[a,b]\to\mathbb{R}\) is strictly monotone. Indeed, suppose \(f\) were neither strictly increasing nor strictly decreasing on \([a,b]\); then there are \(x_1<x_2<x_3\) in \([a,b]\) with \(f(x_2)\) not between \(f(x_1)\) and \(f(x_3)\) — say \(f(x_1)<f(x_3)<f(x_2)\) (the other orderings are symmetric). Applying the intermediate value theorem to \(f\) on \([x_1,x_2]\) at the value \(f(x_3)\) gives some \(x_4\in(x_1,x_2)\) with \(f(x_4)=f(x_3)\); since \(x_4\neq x_3\) this contradicts injectivity. This is exactly the property used earlier to guarantee that a continuous injective real function has an inverse function whose graph is well-defined.
Compact Sets and the Extreme Value Theorem
A set \(K\subseteq\mathbb{R}\) is compact if every sequence in \(K\) has a subsequence converging to a point of \(K\). By Bolzano–Weierstrass, every closed bounded interval \([a,b]\) is compact: any sequence in \([a,b]\) is bounded, hence has a convergent subsequence, and its limit stays in \([a,b]\) since the interval is closed. An unbounded set, or an interval missing an endpoint like \((0,1)\) (take \(x_n=1/n\), whose only possible limit \(0\) is not in the set), is not compact.
Continuous image of a compact set is compact. If \(f:K\to\mathbb{R}\) is continuous and \(K\) is compact, then \(f(K)\) is compact. Given a sequence \(y_n=f(x_n)\) in \(f(K)\), compactness of \(K\) gives a subsequence \(x_{n_k}\to x_0\in K\); continuity then gives \(y_{n_k}=f(x_{n_k})\to f(x_0)\in f(K)\).
Extreme value theorem. A continuous \(f:K\to\mathbb{R}\) on a nonempty compact set \(K\) attains a maximum and a minimum. \(f(K)\) is compact, hence bounded, so \(M:=\sup f(K)\) is finite; picking \(y_n\in f(K)\) with \(y_n\to M\) and using compactness of \(f(K)\) to pass to a subsequence converging to a point of \(f(K)\) forces that point to equal \(M\) (limits are unique), so \(M\in f(K)\), i.e. \(M=f(c)\) for some \(c\in K\). The same argument applied to \(-f\) gives the minimum.
Continuity of the inverse function. If \(f:K\to\mathbb{R}\) is continuous and injective on a compact set \(K\), then \(f^{-1}:f(K)\to K\) is continuous. Let \(y_n\to y_0\) in \(f(K)\), with \(x_n:=f^{-1}(y_n)\) and \(x_0:=f^{-1}(y_0)\); it suffices to show \(x_n\to x_0\). If not, some subsequence stays at distance \(\geq\eta>0\) from \(x_0\); compactness of \(K\) extracts a further subsequence \(x_{n_k}\to x_1\in K\) with \(|x_1-x_0|\geq\eta>0\), so \(x_1\neq x_0\). Continuity of \(f\) gives \(f(x_{n_k})\to f(x_1)\), but also \(f(x_{n_k})=y_{n_k}\to y_0=f(x_0)\), so \(f(x_1)=f(x_0)\) — contradicting injectivity. This generalizes the strictly-monotone-inverse statement above (which needed \([a,b]\) and strict monotonicity) to any compact domain, at the cost of no longer describing the inverse's monotonicity directly.
The Fundamental Theorem of Algebra
Every non-constant polynomial \(p(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_0\) with complex coefficients has a complex root.
Proof. \(|p|:\mathbb{C}\to\mathbb{R}_{\geq0}\) is continuous (composition of the continuous map \(z\mapsto p(z)\), itself built from addition/multiplication, with the continuous absolute value). As in the odd-degree-polynomial argument above, \(|p(z)|\to\infty\) as \(|z|\to\infty\), so there is an \(R>0\) with \(|p(z)|>|p(0)|\) whenever \(|z|>R\); the closed disk \(\overline{D}_R:=\{z:|z|\leq R\}\) is compact (the same Bolzano–Weierstrass argument as for a real interval, applied coordinatewise), so the extreme value theorem gives a \(z_0\in\overline{D}_R\) minimizing \(|p|\) on \(\overline{D}_R\) — and since \(|p(0)|\) is itself at least this minimum while every point outside \(\overline{D}_R\) has strictly larger \(|p|\), \(z_0\) minimizes \(|p|\) over all of \(\mathbb{C}\).
Suppose for contradiction \(p(z_0)\neq0\). Writing \(q(w):=p(z_0+w)/p(z_0)\), \(q\) is a degree-\(n\) polynomial in \(w\) with \(q(0)=1\), and \(|q|\) attains its global minimum \(1\) at \(w=0\). Let \(m\geq1\) be the smallest index with a nonzero coefficient of \(w^m\) in \(q\), so \(q(w)=1+a w^m+O(w^{m+1})\) with \(a\neq0\). Choose \(\beta\in\mathbb{C}\) with \(\beta^m=-1/a\) (an \(m\)-th root exists by the complex roots construction), and consider \(q(t\beta)\) for small real \(t>0\):
\[ q(t\beta)=1+a(t\beta)^m+O(t^{m+1})=1-t^m+O(t^{m+1}), \]
so \(|q(t\beta)|\leq1-t^m+O(t^{m+1})<1\) for all sufficiently small \(t>0\) — contradicting that \(1\) is the global minimum of \(|q|\). Hence \(p(z_0)=0\).
Equivalently, every degree-\(n\) complex polynomial factors into exactly \(n\) linear factors (counting multiplicity): after finding one root \(z_0\), polynomial division gives \(p(z)=(z-z_0)q(z)\) with \(q\) of degree \(n-1\), and induction finishes the factorization. Real polynomials, by contrast, need not factor into real linear factors at all — \(z^2+1\) is exactly the example that motivates the imaginary unit in the first place.
Limits of Functions and Continuous Extension
A point \(a\) is an accumulation point of \(D\) if some sequence in \(D\setminus\{a\}\) converges to \(a\). Recall from the limits chapter that \(\lim_{x\to a}f(x)=L\) means \(f(x_n)\to L\) for every sequence \(x_n\to a\) in \(D\setminus\{a\}\); \(f\) is continuous at \(a\in D\) exactly when \(a\) is not an accumulation point of \(D\), or \(\lim_{x\to a}f(x)=f(a)\).
\(f(x)=\sin(1/x)\) has no limit as \(x\to0\) (it takes every value in \([-1,1]\) infinitely often on any neighborhood of \(0\)), while \(g(x)=x\sin(1/x)\) does have limit \(0\), by the squeeze theorem applied to \(|g(x)|\leq|x|\). This is exactly the kind of function whose removable discontinuity at \(0\) a continuous extension can repair.
Existence of a continuous extension. Let \(a\) be an accumulation point of \(D\) not in \(D\). Then \(f:D\to\mathbb{R}\) extends to a continuous function on \(D\cup\{a\}\) if and only if \(\lim_{x\to a}f(x)\) exists, and the extension is then unique (forced to send \(a\) to that limit).
Proof. If the limit \(L\) exists, define \(\tilde f(a):=L\) and \(\tilde f:=f\) elsewhere; continuity of \(\tilde f\) at \(a\) is exactly the statement \(\tilde f(x_n)\to\tilde f(a)\) for \(x_n\to a\), which is the limit condition (using \(\tilde f=f\) on \(D\)). Conversely, if a continuous extension \(\tilde f\) exists, the sequential criterion at \(a\) gives \(f(x_n)=\tilde f(x_n)\to\tilde f(a)\) for every \(x_n\to a\), so the limit exists and equals \(\tilde f(a)\).
The Cauchy Criterion for Functional Limits
\(\lim_{x\to a}f(x)\) exists if and only if
\[ \forall\epsilon>0\ \exists\delta>0\ \forall x,y\in D\setminus\{a\}:\quad |x-a|<\delta\text{ and }|y-a|<\delta \implies |f(x)-f(y)|<\epsilon, \]
without needing to know the limiting value in advance — exactly the same convenience the Cauchy criterion for sequences provides. If the limit \(L\) exists, the usual \(\epsilon/2\) split gives the condition. Conversely, if the condition holds, take any sequence \(x_n\to a\); it satisfies the Cauchy criterion for sequences (using \(\delta\) to bound how large \(n\) must be), so \(f(x_n)\) converges to some \(L\); and for a second sequence \(x_n'\to a\), interleaving \(x_1,x_1',x_2,x_2',\ldots\) (which still converges to \(a\)) shows \(f(x_n')\) must converge to the same \(L\), so the sequential limit is well-defined independent of the approaching sequence.
Uniform Continuity
\(f:D\to\mathbb{R}\) is uniformly continuous if
\[ \forall\epsilon>0\ \exists\delta>0\ \forall x,y\in D:\quad |x-y|<\delta\implies|f(x)-f(y)|<\epsilon, \]
the same \(\delta\) working simultaneously at every point, rather than only at one fixed point as in plain continuity. Every Lipschitz function is uniformly continuous, with \(\delta=\epsilon/L\) working everywhere at once (the same computation as in Lipschitz \(\Rightarrow\) continuous, just noting \(\delta\) never depended on the point). The converse fails: \(f(x)=1/x\) is continuous on \((0,\infty)\) but not uniformly continuous there. Indeed for \(\epsilon=1\), no \(\delta>0\) works: taking \(x=\delta\) and \(y=\delta/2\) gives \(|x-y|=\delta/2<\delta\), yet \(|f(x)-f(y)|=|1/\delta-2/\delta|=1/\delta\), which exceeds \(1\) once \(\delta<1\).
Uniform continuity extends automatically. If \(f:D\to\mathbb{R}\) is uniformly continuous, then for every accumulation point \(a\) of \(D\) (whether or not \(a\in D\)), \(\lim_{x\to a}f(x)\) exists. Take any \(x_n,x_n'\to a\) in \(D\); given \(\epsilon>0\), pick \(\delta\) from uniform continuity, and note that eventually both \(|x_n-a|<\delta/2\) and \(|x_n'-a|<\delta/2\), so \(|x_n-x_n'|<\delta\) and hence \(|f(x_n)-f(x_n')|<\epsilon\) — in particular this holds for \(x_n'=x_n\) shifted by one index, so \((f(x_n))\) is Cauchy and converges, and the same argument comparing two different approaching sequences shows they converge to the same value. So \(f\) extends continuously to the closure of \(D\) (every accumulation point, together with \(D\) itself).
Continuous on a compact set implies uniformly continuous. If \(f:K\to\mathbb{R}\) is continuous and \(K\) is compact, then \(f\) is uniformly continuous on \(K\). Suppose not: some \(\epsilon_0>0\) has, for every \(n\), points \(x_n,y_n\in K\) with \(|x_n-y_n|<1/n\) yet \(|f(x_n)-f(y_n)|\geq\epsilon_0\). Compactness gives a subsequence \(x_{n_k}\to x_0\in K\); since \(|x_{n_k}-y_{n_k}|\to0\), also \(y_{n_k}\to x_0\). Continuity at \(x_0\) then forces both \(f(x_{n_k})\to f(x_0)\) and \(f(y_{n_k})\to f(x_0)\), so \(|f(x_{n_k})-f(y_{n_k})|\to0\) — contradicting \(|f(x_{n_k})-f(y_{n_k})|\geq\epsilon_0\) for every \(k\). (This is why the extreme value theorem's compactness hypothesis cannot simply be replaced by continuity alone: \(1/x\) on \((0,\infty)\) is continuous but neither uniformly continuous nor bounded.)
Compact Sets Are Exactly the Closed Bounded Ones
\(K\subseteq\mathbb{R}\) is compact if and only if it is closed (contains all its accumulation points) and bounded. Boundedness was used implicitly above (Bolzano–Weierstrass needs a bounded sequence to extract a convergent subsequence): if \(K\) were unbounded, the sequence \(x_n\in K\) with \(|x_n|\geq n\) has no convergent subsequence at all. If \(K\) were bounded but missing an accumulation point \(a\), a sequence in \(K\) converging to \(a\) has every subsequence also converging to \(a\notin K\), so no subsequence converges within \(K\). Conversely, if \(K\) is closed and bounded, any sequence in \(K\) is bounded, so Bolzano–Weierstrass gives a convergent subsequence, and its limit is an accumulation point of \(K\) (or eventually constant), hence lies in \(K\) since \(K\) is closed.
Continuity via Open and Closed Sets
A set \(U\subseteq D\) is open in \(D\) if every point of \(U\) has an entire \(\epsilon\)-neighborhood inside \(U\): for each \(a\in U\), some \(\epsilon>0\) satisfies \(\{x\in D:|x-a|<\epsilon\}\subseteq U\). A set is closed in \(D\) exactly when its complement in \(D\) is open, which matches the accumulation-point description used above: if \(C\subseteq D\) contains all its accumulation points that lie in \(D\), then every \(a\in D\setminus C\) is not an accumulation point of \(C\), so some \(\epsilon\)-neighborhood of \(a\) misses \(C\) entirely, i.e. lies in \(D\setminus C\); conversely a point of \(D\setminus C\) with every neighborhood meeting \(C\) would be an accumulation point of \(C\) outside \(C\), contradicting that \(D\setminus C\) is open.
Topological characterization of continuity. \(f:D\to\mathbb{R}\) is continuous if and only if \(f^{-1}(V)\) is open in \(D\) for every open \(V\subseteq\mathbb{R}\).
Proof. Suppose \(f\) is continuous and \(V\subseteq\mathbb{R}\) is open. For \(a\in f^{-1}(V)\), openness of \(V\) gives \(\epsilon>0\) with \((f(a)-\epsilon,f(a)+\epsilon)\subseteq V\); continuity at \(a\) then gives \(\delta>0\) with \(x\in D,\,|x-a|<\delta\Rightarrow f(x)\in(f(a)-\epsilon,f(a)+\epsilon)\subseteq V\), i.e. \(\{x\in D:|x-a|<\delta\}\subseteq f^{-1}(V)\). So \(f^{-1}(V)\) is open in \(D\). Conversely, suppose every preimage of an open set is open in \(D\). Fix \(a\in D\) and \(\epsilon>0\); the interval \(V:=(f(a)-\epsilon,f(a)+\epsilon)\) is open, so \(f^{-1}(V)\) is open in \(D\) and contains \(a\), giving \(\delta>0\) with \(\{x\in D:|x-a|<\delta\}\subseteq f^{-1}(V)\) — exactly the \(\epsilon\)-\(\delta\) condition for continuity at \(a\).
Taking complements gives the closed-set version for free: \(f\) is continuous if and only if \(f^{-1}(C)\) is closed in \(D\) for every closed \(C\subseteq\mathbb{R}\), since \(f^{-1}(\mathbb{R}\setminus C)=D\setminus f^{-1}(C)\). The real point of this reformulation is that it no longer mentions \(\epsilon\) or \(\delta\) at all — only which sets are open — so the same statement, and the same two-line proof, carries over unchanged to any setting where "open" makes sense, well beyond \(\mathbb{R}\) or \(\mathbb{C}\).
Uniform Convergence of Function Sequences
A sequence of functions \(f_n:D\to\mathbb{R}\) converges pointwise to \(f:D\to\mathbb{R}\) if \(f_n(x)\to f(x)\) for every fixed \(x\in D\) — an ordinary sequence limit at each point separately, possibly with a different rate of convergence at each \(x\). It converges uniformly to \(f\) if
\[ \forall\epsilon>0\ \exists N\ \forall n\geq N\ \forall x\in D:\quad |f_n(x)-f(x)|<\epsilon, \]
the same \(N\) working simultaneously at every point of \(D\) — equivalently, \(\sup_{x\in D}|f_n(x)-f(x)|\to0\). Uniform convergence always implies pointwise convergence (fix \(x\) and drop the "for all \(x\)"), but not conversely.
\(f_n(x):=x^n\) on \(D=[0,1]\) converges pointwise to
\[ f(x)=\begin{cases}0,&0\leq x<1,\\1,&x=1,\end{cases} \]
since \(x^n\to0\) for \(x\in[0,1)\) and \(1^n=1\). Every \(f_n\) is continuous, yet \(f\) is not (it jumps at \(x=1\)), so the convergence cannot be uniform: \(\sup_{x\in[0,1)}|x^n-0|=1\) for every \(n\) (values of \(x^n\) get arbitrarily close to \(1\) as \(x\uparrow1\)), so the sup-distance to \(f\) never shrinks below \(1\).
Uniform limit of continuous functions is continuous. If \(f_n\to f\) uniformly on \(D\) and every \(f_n\) is continuous at \(a\in D\), then \(f\) is continuous at \(a\).
Proof. Given \(\epsilon>0\), uniform convergence gives an \(N\) with \(|f_N(x)-f(x)|<\epsilon/3\) for every \(x\in D\); continuity of \(f_N\) at \(a\) gives \(\delta>0\) with \(|x-a|<\delta\Rightarrow|f_N(x)-f_N(a)|<\epsilon/3\). For \(x\in D\) with \(|x-a|<\delta\), the triangle inequality gives
\[ |f(x)-f(a)|\leq|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\frac\epsilon3+\frac\epsilon3+\frac\epsilon3=\epsilon. \]
The \(x^n\)-on-\([0,1]\) example shows the compactness of \(D\) is irrelevant here and continuity really does need the uniform hypothesis: the same sequence converges pointwise on the same compact set without the limit staying continuous. This is exactly why a convergent power series is continuous on any closed sub-disk strictly inside its radius of convergence — the partial sums are continuous (finite sums of continuous monomials) and converge uniformly there, so the theorem transfers continuity to the limit.
Application: Powers with Real Exponents
The rational-exponent power \(x^r\) is already defined for \(x>0\) and \(r\in\mathbb{Q}\). Extending it to irrational exponents rigorously is a direct application of uniform continuity and continuous extension, rather than a new idea.
Fix \(x>0\) and \(m\in\mathbb{N}\). For rationals \(r,s\in[-m,m]\), writing \(r=s+h\),
\[ |x^r-x^s|=x^s\bigl|x^h-1\bigr|. \]
On \([-m,m]\), \(x^s\) is bounded by \(C:=\max(x^m,x^{-m})\), and a short induction on rational \(h\) in a bounded range shows \(|x^h-1|\leq K|h|\) for a constant \(K\) depending only on \(x\) and \(m\) (the same kind of Bernoulli-inequality bound used earlier for real \(n\)-th roots, applied to \(x^{1/q}\) for the denominator \(q\) of \(h\)). So
\[ |x^r-x^s|\leq CK|r-s|\quad\text{for all }r,s\in\mathbb{Q}\cap[-m,m], \]
i.e. \(r\mapsto x^r\) is Lipschitz, hence uniformly continuous, on \(\mathbb{Q}\cap[-m,m]\). Since \(\mathbb{Q}\) is dense in \(\mathbb{R}\), the closure of \(\mathbb{Q}\cap[-m,m]\) is all of \([-m,m]\), so uniform continuity extends \(x^r\) to a unique continuous function on \([-m,m]\):
\[ x^t:=\lim_{n\to\infty}x^{r_n}\quad\text{for any rational sequence }r_n\to t. \]
Since \(m\) was arbitrary, this defines \(x^t\) for every real \(t\), agreeing with the already-known value whenever \(t\in\mathbb{Q}\). The familiar power laws \(x^{s+t}=x^sx^t\) and \((x^s)^t=x^{st}\) carry over from \(\mathbb{Q}\) to \(\mathbb{R}\) for free: both sides are continuous functions of \((s,t)\) (compositions of continuous multiplication/exponentiation) agreeing on the dense set \(\mathbb{Q}\times\mathbb{Q}\), so they agree everywhere by the sequential criterion applied to rational sequences converging to \(s\) and \(t\).