Continuity and Limits of Functions

A continuous function is one that respects limits, sending nearby inputs to nearby outputs and convergent sequences to convergent sequences. The notion turns the static completeness of the real numbers into theorems about functions, three of which are the working tools of calculus, the intermediate value theorem, the extreme value theorem, and uniform continuity. This post proves them from the least upper bound axiom and the Bolzano-Weierstrass theorem alone, so it builds only on the previous post and underlies all of differential and integral calculus [1], [2].

#Limits of functions

Definition1

Let $f$ be defined on a full punctured neighbourhood $(c-r,c+r)\setminus\{c\}$ for some $r>0$ , so that $c$ is a limit point of $\operatorname{dom}(f)$ , possibly not in it. The limit of $f$ at $c$ is $L$ , written $\lim_{x\to c}f(x)=L$ , when for every $\varepsilon>0$ there is a $\delta>0$ such that $0<\abs{x-c}<\delta$ implies $\abs{f(x)-L}<\varepsilon$ .

The definition mirrors the definition of a sequence limit, with the index going to infinity replaced by the input approaching $c$ . In fact the two notions are equivalent.

Theorem2

$\lim_{x\to c}f(x)=L$ if and only if $f(x_n)\to L$ for every sequence $x_n\to c$ with $x_n\neq c$ .

Proof

Suppose $\lim_{x\to c}f(x)=L$ and $x_n\to c$ with $x_n\neq c$ . Given $\varepsilon>0$ , take $\delta$ from the definition and then $N$ with $\abs{x_n-c}<\delta$ for $n\ge N$ , which holds since $x_n\to c$ . For these $n$ , $0<\abs{x_n-c}<\delta$ , so $\abs{f(x_n)-L}<\varepsilon$ , proving $f(x_n)\to L$ . Conversely, suppose the limit fails. Then some $\varepsilon>0$ admits no working $\delta$ , so for each $n$ the choice $\delta=1/n$ fails, giving a point $x_n\in\operatorname{dom}(f)$ with $0<\abs{x_n-c}<1/n$ yet $\abs{f(x_n)-L}\ge\varepsilon$ ; such an $x_n$ exists because $c$ is a limit point of $\operatorname{dom}(f)$ . Then $x_n\to c$ with $x_n\neq c$ but $f(x_n)\not\to L$ , contradicting the sequential hypothesis.

The sequential characterisation transports every limit law for sequences to functions at one stroke. Sums, products, and quotients of functions with limits have the expected limits, because the corresponding sequences do.

#Continuity

Definition3

A function $f$ is continuous at $c$ when $\lim_{x\to c}f(x)=f(c)$ , that is when for every $\varepsilon>0$ there is a $\delta>0$ with $\abs{x-c}<\delta$ implying $\abs{f(x)-f(c)}<\varepsilon$ . It is continuous on a set when it is continuous at each point.

By Theorem 2 applied with $L=f(c)$ , continuity at $c$ gives $f(x_n)\to f(c)$ for every $x_n\to c$ with $x_n\neq c$ . A general sequence $x_n\to c$ splits into terms with $x_n=c$ , where $f(x_n)=f(c)$ exactly, and terms with $x_n\neq c$ , which converge to $f(c)$ by the punctured statement, so interleaving the two gives $f(x_n)\to f(c)$ with no restriction; conversely the unrestricted criterion contains the punctured one. Hence continuity at $c$ is exactly the statement that $f(x_n)\to f(c)$ whenever $x_n\to c$ , so a continuous function commutes with limits, $f(\lim x_n)=\lim f(x_n)$ . Sums, products, quotients with nonzero denominator, and compositions of continuous functions are continuous, each by the matching sequence law. Polynomials are continuous everywhere, and the elementary functions are continuous on their domains.

#The intermediate value theorem

Continuity forbids a function from skipping a value as it moves between two heights. The proof is the first place the least upper bound axiom acts on a function.

Theorem4

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$ , then $f(c)=y$ for some $c\in[a,b]$ .

Proof

Assume $f(a)<y<f(b)$ , the other case following by negating $f$ . Let $S=\{x\in[a,b]:f(x)<y\}$ , which is nonempty since $a\in S$ and bounded above by $b$ , so $c=\sup S$ exists by the least upper bound axiom. Choosing $x_n\in S$ with $c-1/n<x_n \le c$ , possible because $c-1/n$ is not an upper bound of $S$ , gives a sequence in $S$ with $x_n\to c$ , and since each $f(x_n)<y$ continuity gives $f(c)=\lim f(x_n)\le y$ . If $f(c)<y$ , then since $f(c)<y<f(b)$ we have $c\neq b$ , so $c<b$ . By continuity there is a $\delta>0$ with $f<y$ on $(c-\delta,c+\delta)\cap[a,b]$ , so any point with $c<x<\min(c+\delta,b)$ lies in $S$ , contradicting $c=\sup S$ . Hence $f(c)=y$ , and $c\in[a,b]$ .

The theorem is the reason a continuous function on an interval has an interval as its image, and it is the existence principle behind root-finding, since a continuous function changing sign on $[a,b]$ must vanish in between.

#The extreme value theorem

On a closed bounded interval a continuous function not only stays bounded but reaches its bounds. Here Bolzano-Weierstrass does the work.

Theorem5

A function continuous on $[a,b]$ is bounded and attains a maximum and a minimum.

Proof

For boundedness, suppose $f$ is unbounded above. Then there are $x_n\in[a,b]$ with $f(x_n)\to\infty$ . The sequence $(x_n)$ is bounded, so by the Bolzano-Weierstrass theorem a subsequence $x_{n_k}\to x^\ast\in [a,b]$ , and continuity gives $f(x_{n_k})\to f(x^\ast)$ , a finite number, contradicting $f(x_{n_k})\to \infty$ . So $f$ is bounded above, and bounded below by the same argument on $-f$ . Let $M=\sup_{[a,b]}f$ , finite by boundedness, and take $x_n$ with $f(x_n)\to M$ . A convergent subsequence $x_{n_k}\to x^\ast$ has $f(x^\ast)=\lim f(x_{n_k})=M$ , so the supremum is attained. The minimum is the maximum of $-f$ .

Boundedness fails on an open interval, where $1/x$ runs off to infinity on $(0,1)$ , and the identity function $x\mapsto x$ attains neither bound on $(0,1)$ , so the theorem genuinely needs the interval closed and bounded, the two halves of compactness on the line.

#Uniform continuity

Continuity allows the tolerance $\delta$ to depend on the point. Uniform continuity demands a single $\delta$ that works everywhere, and on a closed bounded interval continuity already delivers it.

Definition6

A function $f$ is uniformly continuous on a set when for every $\varepsilon>0$ there is a single $\delta>0$ such that $\abs{x-y}<\delta$ implies $\abs{f(x)-f(y)}<\varepsilon$ for all $x,y$ in the set.

Theorem7

A function continuous on $[a,b]$ is uniformly continuous there.

Proof

Suppose not. Then some $\varepsilon>0$ resists every $\delta$ , so for each $n$ the choice $\delta=1/n$ fails, giving points $x_n,y_n\in[a,b]$ with $\abs{x_n-y_n}<1/n$ but $\abs{f(x_n)-f(y_n)}\ge\varepsilon$ . By Bolzano-Weierstrass a subsequence $x_{n_k}\to x^\ast\in[a,b]$ , and since $\abs{x_{n_k}-y_{n_k}}\to 0$ the matched subsequence $y_{n_k}\to x^\ast$ too. Continuity gives $f(x_{n_k})\to f(x^\ast)$ and $f(y_{n_k})\to f(x^\ast)$ , so $\abs{f(x_{n_k})-f(y_{n_k})}\to 0$ , contradicting the gap $\ge\varepsilon$ .

Uniform continuity is the strength a function needs for the Riemann integral and for interchange of limits, since it controls oscillation by a length rather than by location. The calculus of a single real variable thus rests on completeness. The next step generalises continuity to metric spaces, where compactness replaces the closed bounded interval and the same proofs run with sequences exchanged for nets of points.

[1]

S. Abbott, Understanding Analysis, 2nd ed. Springer, 2015.

[2]

W. Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{continuity-and-limits,
  author = {Zac Kienzle},
  title  = {Continuity and Limits of Functions},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/continuity-and-limits}
}

#Limits of functions

Definition1

The definition mirrors the definition of a sequence limit, with the index going to infinity replaced by the input approaching $c$ . In fact the two notions are equivalent.

Theorem2

$\lim_{x\to c}f(x)=L$ if and only if $f(x_n)\to L$ for every sequence $x_n\to c$ with $x_n\neq c$ .

Proof

#Continuity

Definition3

#The intermediate value theorem

Continuity forbids a function from skipping a value as it moves between two heights. The proof is the first place the least upper bound axiom acts on a function.

Theorem4

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$ , then $f(c)=y$ for some $c\in[a,b]$ .

Proof

#The extreme value theorem

On a closed bounded interval a continuous function not only stays bounded but reaches its bounds. Here Bolzano-Weierstrass does the work.

Theorem5

A function continuous on $[a,b]$ is bounded and attains a maximum and a minimum.

Proof

#Uniform continuity

Continuity allows the tolerance $\delta$ to depend on the point. Uniform continuity demands a single $\delta$ that works everywhere, and on a closed bounded interval continuity already delivers it.

Definition6

Theorem7

A function continuous on $[a,b]$ is uniformly continuous there.

Proof

[1]

S. Abbott, Understanding Analysis, 2nd ed. Springer, 2015.

[2]

W. Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{continuity-and-limits,
  author = {Zac Kienzle},
  title  = {Continuity and Limits of Functions},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/continuity-and-limits}
}