3.1 Function limits and continuity

Limits and domain points

Definition

Function limit

Let $A\subset\mathbb R$, and let $c$ be an accumulation point of $A$ (every punctured neighbourhood of $c$ meets $A$). For $f:A\to\mathbb R$, we say $\lim_{x\to c,\,x\in A}f(x)=L$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $x\in A$ and $0<|x-c|<\delta$ imply $|f(x)-L|<\varepsilon$.

The value $f(c)$, if defined, is irrelevant to this punctured limit. The accumulation-point condition prevents the definition from being vacuously true for every $L$ at an isolated point. Domain restrictions matter: endpoint limits on a closed interval are one-sided relative to the domain.

Worked example

An epsilon–delta estimate

For $f(x)=x^2$ at $c=2$, target $L=4$. If $|x-2|<1$, then $|x+2|<5$. Thus $|x^2-4|=|x-2||x+2|<5|x-2|$. Given $\varepsilon>0$, choose $\delta=\min(1,\varepsilon/5)$. Then $0<|x-2|<\delta$ gives $|x^2-4|<\varepsilon$.

Theorem

Sequential criterion for a function limit

For an accumulation point $c$ of $A$, the limit of $f$ at $c$ equals $L$ if and only if $f(x_n)\to L$ for every sequence $x_n\in A\setminus\{c\}$ with $x_n\to c$.

Proof

Proving the criterion

Given the epsilon–delta limit and any such sequence, eventually $0<|x_n-c|<\delta$, so $|f(x_n)-L|<\varepsilon$. Conversely, if the limit fails, some $\varepsilon_0>0$ defeats every $\delta>0$. For each $n$, choose $x_n\in A$ with $0<|x_n-c|<1/n$ and $|f(x_n)-L|\ge\varepsilon_0$. Then $x_n\to c$ but $f(x_n)$ does not converge to $L$, contradicting the sequential hypothesis.

Worked example

Two approaches can disprove a limit

For $f(x)=\sin(1/x)$ on $\mathbb R\setminus\{0\}$, the sequences $x_n=(\pi/2+2\pi n)^{-1}$ and $y_n=(3\pi/2+2\pi n)^{-1}$ both approach $0$, but $f(x_n)=1$ and $f(y_n)=-1$. No common limit exists. To prove a limit, by contrast, a few selected paths or sequences do not suffice: the criterion requires every qualifying sequence.

Continuity and the intermediate value theorem

Definition

Continuity relative to a domain

For $c\in A$, $f:A\to\mathbb R$ is continuous at $c$ if for every $\varepsilon>0$ there is $\delta>0$ such that $x\in A$ and $|x-c|<\delta$ imply $|f(x)-f(c)|<\varepsilon$. At an isolated point of $A$ this is automatic. At an accumulation point it is equivalent to the limit equaling $f(c)$.

Theorem

Intermediate value theorem

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

Proof

Completeness locates a crossing

If $y=f(a)$ or $y=f(b)$, take the corresponding endpoint. Otherwise, assume $f(a)<y<f(b)$; the reverse order follows by replacing both $f$ by $-f$ and $y$ by $-y$. The set $S=\{x\in[a,b]:f(x)\le y\}$ is nonempty and bounded above. Let $c=\sup S$. If $f(c)<y$, continuity gives nearby points to the right with value below $y$; since $c<b$, this contradicts the upper-bound property. If $f(c)>y$, continuity gives a neighbourhood of $c$ with value above $y$; the supremum approximation property supplies an element of $S$ in that neighbourhood, a contradiction. Thus $f(c)=y$.

Worked example

Existence without a closed-form root

The polynomial $f(x)=x^3+x-1$ is continuous. Since $f(0)=-1$ and $f(1)=1$, the intermediate value theorem gives a root in $(0,1)$. Further, if $0\le x<y\le1$, then $f(y)-f(x)=(y-x)(y^2+xy+x^2+1)>0$, so this root is unique in that interval. The intermediate value theorem alone gives existence, not uniqueness or an explicit root.

Method comparison

Use an epsilon–delta estimate for a direct limit or continuity proof. Use two approaching sequences with incompatible output limits to disprove existence. Use the intermediate value theorem to establish that a continuous function attains an intermediate value; add a separate argument for uniqueness. Always state the domain and verify continuity on the entire interval.

Practice

1. Prove $\lim_{x\to1}(3x+2)=5$ using epsilon–delta language.
2. Is $f(x)=1$ for $x\ne0$, $f(0)=0$ continuous at $0$?
3. Show $x^3+x-3$ has a root in $(1,2)$.
4. Is every function on the discrete domain $\{1,2,3\}$ continuous relative to that domain?

Full solutions

1. Given $\varepsilon>0$, take $\delta=\varepsilon/3$. Then $0<|x-1|<\delta$ implies $|3x+2-5|=3|x-1|<\varepsilon$.
2. No. The punctured limit is $1$, while $f(0)=0$. For $\varepsilon=1/2$, every neighbourhood contains a nonzero point with output distance $1$ from $f(0)$.
3. The polynomial is continuous and takes values $-1$ at $1$ and $7$ at $2$, so IVT gives a root between them.
4. Yes. Each domain point is isolated; choose a sufficiently small $\delta$ that no other domain point lies within it. Then the only possible $x$ is the point itself, and the output difference is zero.

Synthesis

Function limits control all sufficiently nearby domain points; continuity connects the limit to the actual value. Completeness plus continuity yields intermediate values, but not uniqueness without an additional argument.