5.3 Delta-epsilon limits, limit laws, and continuity

Chapter 5 now shifts from sequences to functions. The new difficulty is that a sequence approaches its limit through the discrete parameter n, while a function argument x can move toward a point a through infinitely many nearby real values from both sides.

This is why the $\delta$ - $\varepsilon$ definition is needed.

From sequence limits to function limits

In the previous chapter, the definition of

\lim_{n\to\infty}x_n=L

said that every small tolerance around $L$ can be achieved by going far enough out in the sequence.

For functions, we want an analogous sentence:

as x approaches a, the function value f(x) approaches $L$ .

But now “going far enough out” no longer makes sense. We need a new way to say that x is close enough to a. That role is played by $\delta$ .

Open intervals and punctured domains

First specify the kind of domains involved.

Definition

Open interval

An open interval is a subset of $R$ of one of the forms

(b,c)=\{x\in R\mid b<x<c\},

(-\infty,c)=\{x\in R\mid x<c\},

(b,\infty)=\{x\in R\mid b<x\},

where $b,c\in R$ and $b<c$ .

To define a limit at a point a, consider functions defined on an open interval with the point a removed. That removal is important: a limit is about what happens near a, not necessarily what happens at a.

The formal delta-epsilon definition

Definition

Limit of a function at a point

Suppose $I$ is an open interval containing $a\in R$ , and let $f:I\setminus\{a\}\to R$ . We say

\lim_{x\to a}f(x)=L

if and only if for every $\varepsilon>0$ , there exists $\delta>0$ such that

0<|x-a|<\delta \implies |f(x)-L|<\varepsilon.

Read the implication carefully:

$0<|x-a|<\delta$ means x is close to a, but not equal to a;
$|f(x)-L|<\varepsilon$ means the function value lands inside the $\varepsilon$ -band around $L$ .

So the definition says:

every requested output accuracy $\varepsilon$ can be guaranteed by forcing the input x into a sufficiently small punctured neighborhood of a.

Common mistake

The limit definition ignores the value at a

The condition $0<|x-a|$ removes the point a itself. So f(a) may be undefined, or may differ from $L$ , while the limit still exists.

First example: $\lim_{x\to 3}(2x+1)=7$

The basic linear test case is the function $f(x)=2x+3$ at $a=1$ .

Worked example

Choosing `\delta=\varepsilon/2`

Let $f(x)=2x+1$ , $a=3$ , and $L=7$ . Then

|f(x)-7| = |(2x+1)-7| = |2x-6| = 2|x-3|.

So in order to force $|f(x)-7|<\varepsilon$ , it is enough to make

|x-3|<\frac{\varepsilon}{2}.

Therefore choose

\delta=\frac{\varepsilon}{2}.

Then whenever $0<|x-3|<\delta$ ,

|f(x)-7|=2|x-3|<2\delta=2\cdot \frac{\varepsilon}{2}=\varepsilon.

Hence

\lim_{x\to 3}(2x+1)=7.

This example shows the standard proof pattern:

start with $|f(x)-L|$ ,
simplify it into an expression involving $|x-a|$ ,
choose $\delta$ to make that expression smaller than $\varepsilon$ .

Two more basic examples

Worked example

Constant functions

If $f(x)=c$ for all x, then

|f(x)-c|=|c-c|=0<\varepsilon

for every $\varepsilon>0$ and every x.

So any positive number can be chosen as $\delta$ , and

\lim_{x\to a}c=c.

Worked example

A function with a hole

Consider

f:(0,\infty)\setminus\{2\}\to R,\qquad f(x)=\frac{x^2-4}{x-2}.

The formula is undefined at $x=2$ , but for $x\ne 2$ it simplifies to

f(x)=x+2.

So near $x=2$ ,

|f(x)-4|=|(x+2)-4|=|x-2|.

Therefore choosing $\delta=\varepsilon$ works. If $0<|x-2|<\delta$ , then

|f(x)-4|<\varepsilon.

Hence

\lim_{x\to 2}\frac{x^2-4}{x-2}=4,

even though f(2) is not defined.

A delta-neighbourhood mapped into an epsilon-band

Figure. The $\delta$ - $\varepsilon$ definition links an input neighborhood of a to an output band around $L$ . The proof task is to choose $\delta$ so that this implication always succeeds.

Explore the definition interactively

The explorer below lets you choose one of the examples, select $\varepsilon$ , and test sample inputs x. The point is to see the implication

0<|x-a|<\delta \implies |f(x)-L|<\varepsilon

as a relationship between an input strip and an output band.

Read and try

Check one delta-epsilon implication geometrically

The explorer links the x-side delta-condition and the y-side epsilon-condition so readers can see the formal implication as two coordinated geometric tests.

0 < |x - a| < δ, with δ = 0.25

x = 2.8

|f(x) - L| < ε, with ε = 0.5

f(x) = 6.6

0 < |x - a| < δ

|2.8 - 3| = 0.2

The chosen x really lies inside the delta-neighbourhood.

|f(x) - L| < ε

|6.6 - 7| = 0.4

The function value lands inside the epsilon-band around L.

How limits can fail to exist

Next, consider how to show that a proposed limit does not exist.

To prove $\lim_{x\to a}f(x)\ne L$ , one must find a single $\varepsilon>0$ such that no matter how small $\delta$ is chosen, some point x with $0<|x-a|<\delta$ still satisfies

|f(x)-L|\ge \varepsilon.

Example 22 uses

f(x)=\frac{|x|}{x},

which equals 1 for $x>0$ and $-1$ for $x<0$ .

Worked example

Why `|x|/x` has no limit at `0`

Take $\varepsilon=1$ . For any $\delta>0$ , choose

x_1=\frac{\delta}{2}>0, \qquad x_2=-\frac{\delta}{2}<0.

Then both satisfy $|x_i|<\delta$ , but

f(x_1)=1,\qquad f(x_2)=-1.

No single real number $L$ can lie within distance 1 of both 1 and $-1$ . Therefore the function has no limit at 0.

This example captures a common failure mode: left-hand and right-hand behavior do not agree.

Algebra of limits

Once some basic limits are known, limit laws show how to build new ones from old ones.

Theorem

Limit laws

If $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$ , then:

$\lim_{x\to a}(f(x)+g(x))=L+M$ ,
$\lim_{x\to a}(f(x)g(x))=LM$ ,
if $M\ne 0$ , then

\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}.

The proof of the sum law is the cleanest illustration: split the allowed error $\varepsilon$ into two halves and use the triangle inequality,

|f(x)+g(x)-(L+M)|\le |f(x)-L|+|g(x)-M|.

The product law needs one extra idea: before estimating the product error, one must first show that |f(x)| stays bounded near a.

Worked example

Using the sum law

Consider $f(x)=2x+1$ and $g(x)=\sqrt{x}$ near $x=4$ . Once the basic limits

\lim_{x\to 4}(2x+1)=9, \qquad \lim_{x\to 4}\sqrt{x}=2

are known, the sum law gives

\lim_{x\to 4}\bigl((2x+1)+\sqrt{x}\bigr)=9+2=11.

The sequential characterization

The sequential criterion connects function limits back to sequence limits.

Theorem

Sequential characterization of function limits

Let $f:I\setminus\{a\}\to R$ . Then

\lim_{x\to a}f(x)=L

if and only if for every sequence $(x_n)$ with $x_n\in I\setminus\{a\}$ for all n and $x_n\to a$ , we have

\lim_{n\to\infty}f(x_n)=L.

This theorem is extremely useful because it gives two complementary methods:

to prove a limit exists, check every sequence approaching a;
to prove a limit does not exist, find two sequences approaching a that give incompatible output behavior.

Example 25 applies this idea to f(x)=\sin(1/x) near 0.

Worked example

Two sequences proving that `\sin(1/x)` has no limit at `0`

Take

x_n=\frac{1}{2\pi n}, \qquad y_n=\frac{1}{2\pi n+\pi/2}.

Then $x_n\to 0$ and $y_n\to 0$ , but

f(x_n)=\sin(2\pi n)=0\to 0,

while

f(y_n)=\sin(2\pi n+\pi/2)=1\to 1.

Since two sequences approaching the same point produce different limiting function values, \lim_{x\to 0}\sin(1/x) does not exist.

Continuity

One of the main purposes of limits is to make the idea of continuity precise.

Definition

Continuity at a point

Let $f:I\to R$ and $a\in I$ . The function f is continuous at a if

\lim_{x\to a}f(x)=f(a).

Equivalently, in $\varepsilon$ - $\delta$ language:

\forall \varepsilon>0\ \exists \delta>0\text{ such that } |x-a|<\delta \implies |f(x)-f(a)|<\varepsilon.

Notice what changed: the condition $0<|x-a|$ disappeared. Continuity really does care about the function value at the point.

A standard discontinuous example is:

f(x)= \begin{cases} 0,& x\ne 0,\\ 1,& x=0. \end{cases}

Here the nearby values are all 0, so the limit near 0 is not $f(0)=1$ . Therefore the function is discontinuous at 0.

Common mistake

A function can have a limit at a without being defined at a

The hole example shows this clearly. Limits are about nearby values. Continuity adds the extra requirement that the function is defined at the point and that its value agrees with the limit.

Quick checks

Quick check

Why does the function-limit definition use $0 < |x-a| < \delta$ instead of only $|x-a| < \delta$ ?

Think about whether the value at a should matter for the limit.

Solution

Answer

Quick check

For the proof that $\lim_{x\to 3}(2x+1)=7$ , what choice of delta works for a given epsilon?

Use the identity $|f(x)-7|=2|x-3|$ .

Solution

Answer

Quick check

Why can $\frac{x^2-4}{x-2}$ have a limit at $x=2$ even though the formula is undefined there?

Recall what the limit definition does and does not inspect.

Solution

Answer

Exercises

Quick check

Prove that $\lim_{x\to 4}\sqrt{x}=2$ .

Rationalize $|\sqrt{x}-2|$ and also keep x near 4 so the denominator stays away from zero.

Solution

Guided solution

Quick check

Use the sequential characterization to show again that `\lim_{x\to 0}\sin(1/x)` does not exist.

You only need two carefully chosen sequences.

Solution

Guided solution

Quick check

What extra condition must be added to ‘the limit of f(x) as x approaches a exists’ in order to conclude that f is continuous at a?

Compare the definitions of limit and continuity.

Solution

Guided solution

Read this after 5.1 Sequences and epsilon-N limits and 5.2 Cauchy sequences and another model of the reals. For the order-theoretic background behind completeness, see 4.3 Completeness and gaps in Q.

5.3 Delta-epsilon limits, limit laws, and continuity

MATH1090: Set theory

From sequence limits to function limits

Open intervals and punctured domains

Open interval

The formal delta-epsilon definition

Limit of a function at a point

The limit definition ignores the value at a

First example: lim⁡x→3(2x+1)=7\lim_{x\to 3}(2x+1)=7limx→3​(2x+1)=7

Choosing \delta=\varepsilon/2

Two more basic examples

Constant functions

A function with a hole

Explore the definition interactively

Check one delta-epsilon implication geometrically

How limits can fail to exist

Why |x|/x has no limit at 0

Algebra of limits

Limit laws

Using the sum law

The sequential characterization

Sequential characterization of function limits

Two sequences proving that \sin(1/x) has no limit at 0

Continuity

Continuity at a point

A function can have a limit at a without being defined at a

Quick checks

Why does the function-limit definition use 0<∣x−a∣<δ0 < |x-a| < \delta0<∣x−a∣<δ instead of only ∣x−a∣<δ|x-a| < \delta∣x−a∣<δ?

Answer

For the proof that lim⁡x→3(2x+1)=7\lim_{x\to 3}(2x+1)=7limx→3​(2x+1)=7, what choice of delta works for a given epsilon?

Answer

Why can x2−4x−2\frac{x^2-4}{x-2}x−2x2−4​ have a limit at x=2x=2x=2 even though the formula is undefined there?

Answer

Exercises

Prove that lim⁡x→4x=2\lim_{x\to 4}\sqrt{x}=2limx→4​x​=2.

Guided solution

Use the sequential characterization to show again that \lim_{x\to 0}\sin(1/x) does not exist.

Guided solution

What extra condition must be added to ‘the limit of f(x) as x approaches a exists’ in order to conclude that f is continuous at a?

Guided solution

Related notes

Prerequisites

Key terms in this unit

More notes in this series

First example: $\lim_{x\to 3}(2x+1)=7$

Choosing `\delta=\varepsilon/2`

Why `|x|/x` has no limit at `0`

Two sequences proving that `\sin(1/x)` has no limit at `0`

Why does the function-limit definition use $0 < |x-a| < \delta$ instead of only $|x-a| < \delta$ ?

For the proof that $\lim_{x\to 3}(2x+1)=7$ , what choice of delta works for a given epsilon?

Why can $\frac{x^2-4}{x-2}$ have a limit at $x=2$ even though the formula is undefined there?

Prove that $\lim_{x\to 4}\sqrt{x}=2$ .

Use the sequential characterization to show again that `\lim_{x\to 0}\sin(1/x)` does not exist.