Chapter 2: ε-δ Arguments (Part 1) — Definitions and Basic Examples

Level: Intermediate

Rigorous definitions of quantifiers, limits, continuity, and sequence limits

Part 1 of a two-part chapter. This chapter is long, so it is split across two files. This article (Part 1) treats the rigorous definitions and basic examples for quantifiers, limits, continuity, and sequence limits. Continue with Chapter 2: ε-δ Arguments (Part 2) — Pitfalls and Extensions.

Why ε-δ Arguments are Needed

The intuitive description "as $x$ approaches $a$, $f(x)$ approaches $L$" is ambiguous.

  • How close is "close enough"?
  • What does "without bound" concretely mean?

ε-δ arguments resolve these ambiguities by using quantifiers.

What are quantifiers

A quantifier is a symbol that specifies the "range" of a variable in a proposition. In mathematics, two main quantifiers are used.

Universal quantifier $\forall$ (for all)

Means "for every ~".

  • $\forall x \in \mathbb{R}, x^2 \geq 0$: "for every real number $x$, $x^2 \geq 0$".
  • $\forall n \in \mathbb{N}, n + 1 > n$: "for every natural number $n$, $n + 1 > n$".

Existential quantifier $\exists$ (there exists)

Means "there exists a ~".

  • $\exists x \in \mathbb{R}, x^2 = 2$: "there exists a real $x$ with $x^2 = 2$" (namely $x = \pm\sqrt{2}$).
  • $\exists n \in \mathbb{N}, n > 100$: "there exists a natural number $n$ greater than $100$".

The order of quantifiers matters

Swapping the order of quantifiers changes the meaning.

  • $\forall \varepsilon > 0, \exists \delta > 0$: "for every $\varepsilon$, there exists a suitable $\delta$ (depending on $\varepsilon$)".
  • $\exists \delta > 0, \forall \varepsilon > 0$: "there exists a single fixed $\delta$ that works for every $\varepsilon$" (a stronger condition).

ε-δ arguments use the former order ($\forall \varepsilon, \exists \delta$).

Pitfalls of the intuitive definition

The verbal description "$f(x) \to L$ as $x \to a$" can lead to wrong conclusions due to its ambiguity.

Problem 1: the oscillating function $\sin(1/x)$

The function $f(x) = \sin(1/x)$ oscillates infinitely many times between $-1$ and $1$ as $x \to 0$.

Intuitively one might say "as $x$ approaches $0$, the value of $f(x)$ fluctuates," but which value it "approaches" is unclear.

Analysis via ε-δ: for any candidate value $L$, take $\varepsilon = 0.1$. No matter how small $\delta$ is chosen, there is some point with $|x| < \delta$ where $\sin(1/x) > L + 0.1$. Hence no limit exists.

Problem 2: the unbounded function $1/x$

The function $f(x) = 1/x$ tends to $+\infty$ and $-\infty$ from the two sides of $0$.

Intuitively one might say "the value gets large," but the direction differs depending on whether $x$ approaches from the left or the right.

Analysis via ε-δ: suppose $\displaystyle \lim_{x \to 0} \dfrac{1}{x} = L$ exists. Then there should be a $\delta$ for which $|1/x - L| < 1$ on $0 < |x| < \delta$. But the sign of $1/x$ differs depending on whether $x$ approaches $0$ from the left or from the right, yielding a contradiction.

Advantages of the ε-δ formulation

By giving the definition rigorously in formula form, the following become possible:

  • Eliminating ambiguity: "approaches" is defined by a precise inequality.
  • Constructing counterexamples: one can logically prove non-existence of a limit.
  • Uniqueness of the limit: a function cannot simultaneously approach two different values.
  • Extension to other concepts: continuity, differentiation, integration, and more complex notions are defined this way.

The Definition of Limit

Limit of a function (ε-δ definition)

$\displaystyle \lim_{x \to a} f(x) = L$ means that the following holds:

$$\forall \varepsilon > 0, \exists \delta > 0, \forall x, (0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon)$$

How to read the definition

  1. For every $\varepsilon > 0$: however small a tolerance is demanded,
  2. there exists a $\delta > 0$: one can choose an appropriate "closeness to $a$" such that
  3. whenever $0 < |x - a| < \delta$: i.e. $x$ is sufficiently close to $a$ but not equal to $a$,
  4. we have $|f(x) - L| < \varepsilon$: $f(x)$ lies within the tolerance band $\pm\varepsilon$ around $L$.
x y a L a−δ a+δ L + ε L - ε

If the orange band (the $a \pm \delta$ range) can be chosen so that the graph stays inside the blue band (the $L \pm \varepsilon$ range), the limit exists.

Cases where the limit does not exist

In the following cases, the $\varepsilon$ condition cannot be satisfied however small $\delta$ is taken, so the limit does not exist.

x y a

Discontinuity
left limit $\neq$ right limit

x y 0

Oscillation
$\sin(1/x)$ oscillates as $x \to 0$

x y 0

Divergence
$1/x$ tends to $\pm\infty$ as $x \to 0$

A Basic Example of ε-δ Proof

Template for an ε-δ proof
  1. Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
  2. Step 2:Scratch workTransform $|f(x) - L| < \varepsilon$ into a bound on $|x - a|$.
  3. Step 3:Choose δDefine $\delta$ as an expression in $\varepsilon$.
  4. Step 4:VerifyCheck that $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$.
  5. Step 5:Conclusion"Therefore the limit is $L$."

Example 1: limit of a linear function

Claim: prove that $\displaystyle \lim_{x \to 2} (3x + 1) = 7$.

Proof

Step 1: Take an arbitrary $\varepsilon > 0$.

Step 2: Determine $\delta$ by working backward from $|f(x) - L| < \varepsilon$.

\begin{align} |f(x) - L| &= |(3x + 1) - 7| \\ &= |3x - 6| \\ &= 3|x - 2| \\ &< \varepsilon \end{align}

This is equivalent to $|x - 2| < \dfrac{\varepsilon}{3}$.

Step 3: Set $\delta = \dfrac{\varepsilon}{3}$.

Step 4: Verify. When $0 < |x - 2| < \delta = \dfrac{\varepsilon}{3}$,

\begin{align} |(3x + 1) - 7| &= |3x - 6| \\ &= 3|x - 2| \\ &< 3 \cdot \dfrac{\varepsilon}{3} \\ &= \varepsilon \end{align}

Step 5: Hence for every $\varepsilon > 0$ the choice $\delta = \dfrac{\varepsilon}{3}$ works.

Therefore $\displaystyle \lim_{x \to 2} (3x + 1) = 7$. $\square$

A More Complex Example: a Quadratic Function

Example 2: limit of a quadratic function

Claim: prove that $\displaystyle \lim_{x \to 3} x^2 = 9$.

Proof

Scratch work (estimate $|x^2 - 9|$ first):

\begin{align} |x^2 - 9| &= |x + 3||x - 3| \end{align}

The factor $|x - 3|$ is bounded by $\delta$, but $|x + 3|$ also needs a bound.

Thought process: why bound $|x+3|$?

Step 1: recognize the product form. $|x^2 - 9| = |x+3||x-3|$ has two factors.

Step 2: decide which to bound with $\delta$. $|x-3|$ is bounded by $\delta$ automatically from the definition, while $|x+3|$ is independent.

Step 3: bound $|x+3|$ by a constant. Following the intuition that "if $|x-3|$ is small, then $|x+3|$ should also be controlled," impose an upper cap on $\delta$ (e.g. $\delta \leq 1$).

Step 4: compute concretely. Assuming $|x-3| < 1$, compute an upper bound for $|x+3|$.

This "assumption" is later realized in the form $\delta = \min(1, \varepsilon/7)$.

Assuming $|x - 3| < 1$ (which is ensured later by $\delta \leq 1$),

\begin{align} |x - 3| &< 1 \\ -1 < x - 3 &< 1 \\ 2 < x &< 4 \\ 5 < x + 3 &< 7 \\ |x + 3| &< 7 \end{align}

Therefore $|x^2 - 9| = |x + 3||x - 3| < 7|x - 3|$.

The proof

Step 1: Take an arbitrary $\varepsilon > 0$.

Step 2: Choose $\delta$. From the scratch work we know $|x^2 - 9| < 7|x - 3|$, so combine the two conditions below.

Choosing δ: why $\min(1, \varepsilon/7)$?

Condition 1: ensure $|x+3| < 7$. To realize the scratch-work assumption $|x-3| < 1$, take $\delta \leq 1$.

Condition 2: ensure $|x^2 - 9| < \varepsilon$

  • The scratch work showed $|x^2 - 9| < 7|x - 3|$.
  • When $|x - 3| < \delta$, we have $|x^2 - 9| < 7\delta$.
  • To make this less than $\varepsilon$, require $7\delta < \varepsilon$, i.e. $\delta < \varepsilon/7$.

Satisfying both conditions. The two conditions $\delta \leq 1$ and $\delta \leq \varepsilon/7$ are simultaneously satisfied by choosing the smaller:

$\delta = \min\left(1, \dfrac{\varepsilon}{7}\right)$

Step 3: When $0 < |x - 3| < \delta$, both conditions from the info box above hold:

  • $\delta \leq 1$ gives $|x - 3| < 1$, so the scratch work gives $|x + 3| < 7$.
  • $\delta \leq \dfrac{\varepsilon}{7}$ gives $|x - 3| < \dfrac{\varepsilon}{7}$.

Step 4: Hence

\begin{align} |x^2 - 9| &= |x + 3||x - 3| \\ &< 7 \cdot \dfrac{\varepsilon}{7} \\ &= \varepsilon \end{align}

Therefore $\displaystyle \lim_{x \to 3} x^2 = 9$. $\square$

The Definition of Continuity

Continuity at a point (ε-δ definition)

A function $f$ is said to be continuous at the point $a$ if the following holds:

$$\forall \varepsilon > 0, \exists \delta > 0, \forall x, (|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon)$$

Notice the differences from the limit definition.

The Definition of Limit
$\forall \varepsilon > 0, \exists \delta > 0, \forall x, ($$0 < |x - a| < \delta$$ \Rightarrow |f(x) - $$L$$| < \varepsilon)$
$x = a$ excluded
↓
$x = a$ included
$L$
↓
$f(a)$
Continuity definition
$\forall \varepsilon > 0, \exists \delta > 0, \forall x, ($$|x - a| < \delta$$ \Rightarrow |f(x) - $$f(a)$$| < \varepsilon)$
Limit: 0 < |x-a| < δ a−δ a a+δ x = a is excluded Continuous: |x-a| < δ a−δ a a+δ x = a is included
Template for a continuity proof
  1. Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
  2. Step 2:Scratch workTransform $|f(x) - f(a)| < \varepsilon$ into a bound on $|x - a|$.
  3. Step 3:Choose δDefine $\delta$ as an expression in $\varepsilon$.
  4. Step 4:VerifyCheck that $|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon$.
  5. Step 5:Conclusion"Therefore $f$ is continuous at $a$."

Example 3: a continuity proof

Claim: $f(x) = x^2$ is continuous at $x = 2$.

Proof

Step 1: Take an arbitrary $\varepsilon > 0$.

Step 2: Set $\delta = \min\left(1, \dfrac{\varepsilon}{5}\right)$.

Step 3: Suppose $|x - 2| < \delta$.

From $\delta \leq 1$ we have $|x - 2| < 1$, i.e. $1 < x < 3$.

Hence $3 < x + 2 < 5$, i.e. $|x + 2| < 5$.

Step 4:

\begin{align} |f(x) - f(2)| &= |x^2 - 4| \\ &= |x + 2||x - 2| \\ &< 5 \cdot \dfrac{\varepsilon}{5} \\ &= \varepsilon \end{align}

Step 5: Therefore $f(x) = x^2$ is continuous at $x = 2$. $\square$

Limits of Sequences

Limit of a sequence (ε-N definition)

$\displaystyle \lim_{n \to \infty} a_n = L$ means that the following holds:

$$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, (n > N \Rightarrow |a_n - L| < \varepsilon)$$

Correspondence between ε-δ and ε-N

The ε-N definition is essentially a "discretized version" of the ε-δ definition. The structural correspondence is summarized below.

Concept Function limit (ε-δ) Sequence limit (ε-N)
Object function $f(x)$ (continuous domain) sequence $\{a_n\}$ (discrete domain)
Variable real $x$ (approaches $a$) natural number $n$ (tends to $\infty$)
Measure of closeness $\delta$: measures the closeness of $x$ to $a$ $N$: measures how far $n$ has progressed
Error tolerance $\varepsilon$: tolerance for $f(x)$ vs. $L$ $\varepsilon$: tolerance for $a_n$ vs. $L$ (same as above)
Condition $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$ $n > N \Rightarrow |a_n - L| < \varepsilon$
Quantifier order $\forall \varepsilon, \exists \delta, \forall x$ $\forall \varepsilon, \exists N, \forall n$

Why ε-N is easier than ε-δ

Sequence limits often feel more intuitive than function limits. Reasons:

  • From one dimension to zero dimensions: $\delta$ is a notion of distance, while $N$ is just an ordinal index.
  • No need to worry about continuity: a sequence takes discrete values, so intermediate-value considerations do not arise.
  • Inverse manipulation is simpler: finding $N$ such that $n > N$ is simpler than finding $\delta$ such that $|x-a| < \delta$.
Template for a sequence-limit proof
  1. Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
  2. Step 2:Scratch workTransform $|a_n - L| < \varepsilon$ into a condition on $n$.
  3. Step 3:Choose NPick a natural number $N > \dfrac{\text{(constant)}}{\varepsilon}$.
  4. Step 4:VerifyCheck that $n > N \Rightarrow |a_n - L| < \varepsilon$.
  5. Step 5:Conclusion"Therefore $\displaystyle\lim_{n \to \infty} a_n = L$."

Example 4: a sequence limit

Claim: prove $\displaystyle \lim_{n \to \infty} \dfrac{1}{n} = 0$.

Proof

Step 1: Take an arbitrary $\varepsilon > 0$.

Step 2: By the Archimedean property, a natural number $N > \dfrac{1}{\varepsilon}$ exists.

Step 3: For $n > N$,

\begin{align} n &> N > \dfrac{1}{\varepsilon} \\ \dfrac{1}{n} &< \dfrac{1}{N} < \varepsilon \end{align}

Step 4: Since $\dfrac{1}{n} > 0$,

\begin{align} \left|\dfrac{1}{n} - 0\right| = \dfrac{1}{n} < \varepsilon \end{align}

Step 5: Therefore $\displaystyle \lim_{n \to \infty} \dfrac{1}{n} = 0$. $\square$

Example 5: a more complex sequence

Claim: prove $\displaystyle \lim_{n \to \infty} \dfrac{2n + 1}{n + 3} = 2$.

Proof

Scratch work

\begin{align} \left|\dfrac{2n + 1}{n + 3} - 2\right| &= \left|\dfrac{2n + 1 - 2(n + 3)}{n + 3}\right| \\ &= \left|\dfrac{2n + 1 - 2n - 6}{n + 3}\right| \\ &= \left|\dfrac{-5}{n + 3}\right| \\ &= \dfrac{5}{n + 3} \end{align}

For $n \geq 1$, $n + 3 > n$, so $\dfrac{5}{n + 3} < \dfrac{5}{n}$.

The proof

Step 1: Take an arbitrary $\varepsilon > 0$.

Step 2: Choose a natural number $N > \dfrac{5}{\varepsilon}$.

Step 3: For $n > N$,

\begin{align} \left|\dfrac{2n + 1}{n + 3} - 2\right| &= \dfrac{5}{n + 3} \\ &< \dfrac{5}{n} \\ &< \dfrac{5}{N} \\ &< \dfrac{5}{\dfrac{5}{\varepsilon}} \\ &= \varepsilon \end{align}

Step 4: Therefore $\displaystyle \lim_{n \to \infty} \dfrac{2n + 1}{n + 3} = 2$. $\square$