What are common mistakes in ε-δ arguments?

Five typical mistakes are: (1) swapping the order of the ε and δ quantifiers, (2) making δ a fixed constant instead of a function of ε, (3) confusing the direction of implication between the scratch work and the actual proof, (4) confusing ⇔ with ⇒, and (5) dropping the condition "0 < |x - a|" (which collapses the distinction between limit and continuity).

How does one prove that a limit does not exist via ε-δ?

By contradiction. (1) Assume lim_{x→a} f(x) = L exists. (2) Choose a concrete ε₀ > 0 that violates the ε-δ definition (typically ε₀ = 1/2). (3) For every δ > 0, construct an x with 0 < |x - a| < δ but |f(x) - L| ≥ ε₀ to derive a contradiction. For an oscillating function like sin(1/x), one alternates between points where the value is ±1.

What is the difference between continuity and uniform continuity?

Ordinary continuity allows a different δ(a, ε) at each point a, whereas uniform continuity requires a single δ(ε) that works at every point. By the Heine-Cantor theorem, a continuous function on a closed interval [a, b] is automatically uniformly continuous. By contrast, 1/x is continuous but not uniformly continuous on (0, 1), because the required δ shrinks as x approaches 0.

Chapter 2: ε-δ Arguments (Part 2) — Pitfalls and Extensions

Q: How does one prove that a limit does not exist via ε-δ?

By contradiction. (1) Assume lim_{x→a} f(x) = L exists. (2) Choose a concrete ε₀ > 0 that violates the ε-δ definition (typically ε₀ = 1/2). (3) For every δ > 0, construct an x with 0 < |x - a| < δ but |f(x) - L| ≥ ε₀ to derive a contradiction. For an oscillating function like sin(1/x), one alternates between points where the value is ±1.

Q: What is the difference between continuity and uniform continuity?

Ordinary continuity allows a different δ(a, ε) at each point a, whereas uniform continuity requires a single δ(ε) that works at every point. By the Heine-Cantor theorem, a continuous function on a closed interval [a, b] is automatically uniformly continuous. By contrast, 1/x is continuous but not uniformly continuous on (0, 1), because the required δ shrinks as x approaches 0.

Level: Intermediate

Common Mistakes, proofs of non-existence, template comparison, uniform continuity, and Exercises

Part 2 of a two-part chapter. This article assumes the definitions and basic examples from Part 1: Chapter 2: ε-δ Arguments (Part 1) — Definitions and Basic Examples. Part 2 covers pitfalls, proofs of non-existence, template comparison, uniform continuity, and exercises.

Common Mistakes

Mistake 1: swapping the order of ε and δ

Wrong: "take $\delta > 0$, and there exists some $\varepsilon > 0$ such that..."

Correct: "for every $\varepsilon > 0$, there exists a $\delta > 0$ such that..."

$\varepsilon$ is given first, and $\delta$ is chosen in response to it.

Mistake 2: not making δ depend on ε

Wrong: "let $\delta = 0.001$" (a fixed value independent of $\varepsilon$).

Correct: $\delta$ is normally a function of $\varepsilon$ (e.g. $\delta = \dfrac{\varepsilon}{3}$).

Mistake 3: confusing the scratch work with the proof

In the scratch work one assumes "$|f(x) - L| < \varepsilon$" and works backward to find $\delta$. In the actual proof one assumes "$|x - a| < \delta$" and derives "$|f(x) - L| < \varepsilon$".

Mind the direction of the logical implication.

Mistake 4: confusing ⇔ with ⇒

Wrong: "$|x - a| < \delta \Leftrightarrow |f(x) - L| < \varepsilon$"

Correct: "$|x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$"

The ε-δ definition requires only the implication in one direction. The reverse direction ($|f(x) - L| < \varepsilon \Rightarrow |x - a| < \delta$) does not hold in general.

Example: for $f(x) = x$, $a = 0$, $L = 0$ one may take $\delta = 2\varepsilon$, but $|f(x)| < \varepsilon$ does not imply $|x| < 2\varepsilon$.

Mistake 5: ignoring the meaning of "$0 < |x - a|$"

Incorrect formulation: "$|x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$" (omitting the "$0 <$" part).

Correct: "$0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$"

The condition "$0 < |x - a|$" means $x \neq a$ and is essential to the definition of a limit.

Reason:

$f$ may not be defined at $x = a$ (e.g. $f(x) = \dfrac{\sin x}{x}$ is undefined at $x = 0$).
The value of $f$ at $x = a$ may differ from the limit value (whereas continuity does involve $f(a)$).
The limit depends only on how $x$ approaches $a$, not on the value at $x = a$.

Contrast: in the definition of continuity one writes "$|x - a| < \delta$" without "$0 <$" precisely because $f(a)$ must equal the limit value.

Proving Non-Existence of a Limit

The ε-δ argument can be used not only to prove directly that a limit exists, but also (via contradiction) that it does not exist.

To prove $\displaystyle \lim_{x \to a} f(x) \neq L$, use proof by contradiction:

Template for a proof by contradiction

Step 1:AssumeAssume that $\displaystyle \lim_{x \to a} f(x) = L$.
Step 2:Choose εFind a concrete $\varepsilon_0 > 0$ that violates the ε-δ definition.
Step 3:Arbitrary δShow that the following holds for every $\delta > 0$.
Step 4:Construct counterexampleExhibit an $x$ with $0 < |x - a| < \delta$ but $|f(x) - L| \geq \varepsilon_0$.
Step 5:Contradict and concludeThe contradiction with the ε-δ definition shows that no value of $L$ works; hence $\displaystyle \lim_{x \to a} f(x)$ does not exist.

Key point: it is necessary to show that every candidate $L$ fails. A single $\varepsilon_0$ is used to derive a contradiction.

Concrete example: $\sin(1/x)$ has no limit as $x \to 0$

Claim

For $f(x) = \sin(1/x)$, the limit $\displaystyle \lim_{x \to 0} \sin(1/x)$ does not exist.

Proof (by contradiction)

Step 1: Assume that $\displaystyle \lim_{x \to 0} \sin(1/x) = L$ exists.

Step 2: Choose $\varepsilon_0 = \dfrac{1}{2}$.

Step 3: By the ε-δ definition, for this $\varepsilon_0$ there should exist a $\delta > 0$ such that $0 < |x| < \delta$ implies $|\sin(1/x) - L| < \dfrac{1}{2}$.

Step 4 (constructing the counterexample):

Within $0 < |x| < \delta$ there exist points where $\sin(1/x) = 1$ and points where $\sin(1/x) = -1$.

When $1/x = \pi/2 + 2\pi k$ (integer $k$), $\sin(1/x) = 1$.
When $1/x = -\pi/2 + 2\pi k$ (integer $k$), $\sin(1/x) = -1$.

Step 5 (deriving the contradiction):

At points with $\sin(1/x) = 1$, $|\sin(1/x) - L| = |1 - L|$.
At points with $\sin(1/x) = -1$, $|\sin(1/x) - L| = |-1 - L|$.

These two quantities cannot both be less than $\dfrac{1}{2}$ simultaneously.

For instance, if $|1 - L| < \dfrac{1}{2}$ then $\dfrac{1}{2} < L < \dfrac{3}{2}$.

Then $|-1 - L| > |{-1 - \dfrac{3}{2}}| = \dfrac{5}{2} > \dfrac{1}{2}$, a contradiction.

Step 6: Therefore the assumption is false, and $\displaystyle \lim_{x \to 0} \sin(1/x)$ does not exist. $\square$

Tips for proving non-existence

Use oscillating functions: functions like $\sin(1/x)$ that oscillate between several values make non-existence easy to demonstrate by contradiction.
Mismatched one-sided limits: e.g. $\displaystyle \lim_{x \to 0^-} 1/x = -\infty$ vs. $\displaystyle \lim_{x \to 0^+} 1/x = +\infty$.
Use discontinuities: step functions and the like have points where no limit exists.
Choosing ε is the key: the heart of the proof is finding an $\varepsilon_0$ for which no value of $L$ works.

Comparison of Proof Templates

The four proof templates — ε-δ for function limits, ε-δ for continuity, ε-N for sequence limits, and proof by contradiction for non-existence — are compared side by side.

ε-δ (function limit)

Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
Step 2:Scratch workTransform $|f(x) - L|$.
Step 3:Choose δDefine $\delta$.
Step 4:Verify$|x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$
Step 5:ConclusionThe limit holds.

Key point: δ-neighborhood excluding $x = a$

Continuity (ε-δ definition)

Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
Step 2:Scratch workTransform $|f(x) - f(a)|$.
Step 3:Choose δDefine $\delta$.
Step 4:Verify$|x-a|<\delta$ ⟹ $|f(x)-f(a)|<\varepsilon$
Step 5:ConclusionContinuity holds.

Key point: δ-neighborhood including $x = a$

Sequence limit (ε-N definition)

Step 1:Setup"Take an arbitrary $\varepsilon > 0$."
Step 2:Scratch workTransform $|a_n - L|$.
Step 3:Choose N$N > \dfrac{c}{\varepsilon}$.
Step 4:Verify$n > N \Rightarrow |a_n - L| < \varepsilon$
Step 5:ConclusionThe sequence converges.

Key point: argument for all $n$ greater than $N$

Proof by Contradiction (Non-Existence of a Limit)

Step 1:AssumeAssume $\displaystyle \lim_{x \to a} f(x) = L$.
Step 2:Choose εFind $\varepsilon_0 > 0$ that violates the ε-δ definition.
Step 3:Arbitrary δFor every $\delta > 0$, construct a counterexample.
Step 4:CounterexampleExhibit $x$ with $0 < |x - a| < \delta$ but $|f(x) - L| \geq \varepsilon_0$.
Step 5:Contradict and concludeThe contradiction with the ε-δ definition shows $\displaystyle \lim_{x \to a} f(x)$ does not exist.

Key point: show that every candidate $L$ leads to a contradiction

Extension: Bridge to Uniform Continuity

What is uniform continuity

In the ε-δ definition of continuity studied so far, δ depends on the point $a$:

∀ε > 0, ∃δ(a, ε) > 0 : |x - a| < δ ⇒ |f(x) - f(a)| < ε

By contrast, in uniform continuity, δ is common to all points:

∀ε > 0, ∃δ(ε) > 0 : ∀x, a : |x - a| < δ ⇒ |f(x) - f(a)| < ε

Key difference: ordinary continuity allows a different δ at each point, while uniform continuity requires "a single δ that works at every point."

When does uniform continuity matter?

Continuous functions on a closed interval: by the Heine-Cantor theorem, a function continuous on a closed interval $[a, b]$ is automatically uniformly continuous.
Definition of the integral: uniform continuity is essential in proving the existence of the Riemann integral.
Numerical computation: uniform continuity allows one to design numerical algorithms with guaranteed accuracy.
Well-behaved functions: uniformly continuous functions have the property that the values at distinct points cannot drift far apart.

Example: $f(x) = x$ is uniformly continuous on $[0, 1]$ (details)

Sketch of proof:

For any $\varepsilon > 0$, take $\delta = \varepsilon$. Then,

|x - a| < δ = ε ⇒ |f(x) - f(a)| = |x - a| < ε

This $\delta = \varepsilon$ works uniformly for every point $a \in [0, 1]$, so the function is uniformly continuous.

Counterexample: $f(x) = 1/x$ is not uniformly continuous on $(0, 1)$ (details)

Sketch of proof: take $\varepsilon = 1$.

For any $\delta > 0$, near the point $a = \delta/2$,

In other words, as $x$ approaches $0$ one must shrink $\delta$, so no single $\delta$ works at every point.

Next steps

Uniform continuity is an important concept in analysis and opens the path to integration theory and functional analysis. At this stage it suffices to know "such a concept exists as an application of the ε-δ definition." The details are studied in courses on real analysis.

Exercises

Problem 1

Prove $\displaystyle \lim_{x \to 1} (2x - 3) = -1$ using the ε-δ argument.

💡 Hint

Step 1: identify $f(x) = 2x - 3$, $a = 1$, $L = -1$.

Step 2 (scratch work): notice that $|(2x-3) - (-1)| = |2x - 2|$. Can this be expressed in terms of $|x - 1|$?

Step 3 (choose δ): if $|x - 1| < \delta$, how is $|2x - 2| = 2|x-1|$ bounded?

Step 4 (relating δ and ε): to ensure $2|x-1| < 2\delta \leq \varepsilon$, how should $\delta$ be chosen?

Problem 2

Prove $\displaystyle \lim_{x \to 0} x^3 = 0$ using the ε-δ argument.

💡 Hint

Step 1: identify $f(x) = x^3$, $a = 0$, $L = 0$.

Step 2 (scratch work): note that $|x^3 - 0| = |x^3| = |x|^3$.

Step 3 (choose δ): to ensure $|x|^3 < \varepsilon$, how small must $|x|$ be? (Aim for a form like $|x| < \sqrt[3]{\varepsilon}$.)

Step 4 (handling complications): unlike Example 2 (the quadratic), this problem is handled directly by $\delta = \sqrt[3]{\varepsilon}$. Think about whether a $\min(\cdot)$ is needed.

Problem 3

Prove $\displaystyle \lim_{n \to \infty} \dfrac{n}{n^2 + 1} = 0$ using the ε-N argument.

💡 Hint

Step 1: identify $a_n = \dfrac{n}{n^2+1}$ and $L = 0$.

Step 2 (scratch work): $\left|\dfrac{n}{n^2+1} - 0\right| = \dfrac{n}{n^2+1}$. Since $n^2 + 1 > n^2$, this is bounded by $\dfrac{n}{n^2} = \dfrac{1}{n}$.

Step 3 (choose N): to ensure $\dfrac{1}{n} < \varepsilon$, it suffices to take $n > \dfrac{1}{\varepsilon}$.

Step 4 (Archimedean property): we are using "for every $\varepsilon > 0$ there exists a natural number $N > \dfrac{1}{\varepsilon}$." This is taken as standard.

Show solutions

Solution to Problem 1

Take an arbitrary $\varepsilon > 0$. Set $\delta = \dfrac{\varepsilon}{2}$.

When $0 < |x - 1| < \delta$,

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

$\square$

Solution to Problem 2

Take an arbitrary $\varepsilon > 0$. Set $\delta = \min(1, \sqrt[3]{\varepsilon})$.

When $0 < |x - 0| < \delta$, $|x| < 1$ and $|x| < \sqrt[3]{\varepsilon}$, so

\begin{align} |x^3 - 0| &= |x|^3 \\ &< (\sqrt[3]{\varepsilon})^3 \\ &= \varepsilon \end{align}

$\square$

Solution to Problem 3

Take an arbitrary $\varepsilon > 0$. Pick $N > \dfrac{1}{\varepsilon}$.

When $n > N$, $n^2 + 1 > n^2 > n$, so

\begin{align} \left|\dfrac{n}{n^2 + 1} - 0\right| &= \dfrac{n}{n^2 + 1} \\ &< \dfrac{n}{n^2} \\ &= \dfrac{1}{n} \\ &< \dfrac{1}{N} \\ &< \varepsilon \end{align}

$\square$