The Rigorous (Epsilon-Delta) Definition of a Limit

Every theorem of calculus you will meet in this course — continuity, the derivative, the integral, convergence of series — rests on a single precise idea: the limit. In secondary school the limit is usually described informally ("$f(x)$ gets close to $L$ as $x$ gets close to $a$"), and that description is good enough to compute with but useless for proving anything, because "gets close to" is not a mathematical statement. This chapter replaces the informal picture with the $\varepsilon$-$\delta$ definition due to Cauchy and Weierstrass, which turns "close" into an inequality and "limit" into a statement with quantifiers that can be verified or refuted.

Have to hand: the absolute value $|x|$ and its triangle inequality $|a+b|\le |a|+|b|$, interval notation, and familiarity with solving inequalities in one variable. Nothing beyond that is assumed.

Why bother with quantifiers. A statement like "$f(x)$ is close to $L$ when $x$ is close to $a$" has two hidden roles reversed if read carelessly. The challenge is how close to $L$ you must get (that is $\varepsilon$); the response is how close to $a$ that forces you to be (that is $\delta$). Getting the order of quantifiers backwards is the single most common error in this entire course.

1. The Definition

Definition (limit of a function). Let $f$ be a function defined on some set containing an open interval around $a$, except possibly at $a$ itself. We say the limit of $f(x)$ as $x$ approaches $a$ is $L$, written

if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x$,

Read this sentence left to right: first an adversary picks an arbitrarily small tolerance $\varepsilon$ on the output; then, knowing $\varepsilon$, you must produce a tolerance $\delta$ on the input that guarantees the output lands within $\varepsilon$ of $L$. The condition $0 < |x-a|$ deliberately excludes $x = a$: the limit describes the behaviour of $f$ near $a$, not at $a$, so $f$ need not even be defined at $a$.

Definition vs. intuition. "$f(x)\to L$ as $x\to a$" is intuition; the $\varepsilon$-$\delta$ inequality chain above is the definition. Every proof in this chapter manipulates the inequality, never the intuition directly — though the intuition is exactly what tells you how to find $\delta$.

2. Finding $\delta$ from $\varepsilon$: A Worked Example

Worked example. Prove that $\lim_{x\to 3}(2x+1) = 7$.

We must show: for every $\varepsilon>0$ there is a $\delta>0$ such that $0<|x-3|<\delta$ implies $|(2x+1)-7|<\varepsilon$.

Scratch work (not part of the proof, but how we find $\delta$). We want $|(2x+1)-7| = |2x-6| = 2|x-3| < \varepsilon$, i.e. $|x-3| < \varepsilon/2$. This suggests $\delta = \varepsilon/2$.

Proof. Let $\varepsilon > 0$ be given. Choose $\delta = \varepsilon/2$. Suppose $0 < |x-3| < \delta$. Then

Since $\varepsilon>0$ was arbitrary, this proves $\lim_{x\to 3}(2x+1)=7$. $\blacksquare$

Notice the logical shape: the scratch work runs backwards (from the desired conclusion to a condition on $\delta$), but the proof itself runs forwards (assume $|x-3|<\delta$, derive $|f(x)-L|<\varepsilon$). A proof that only shows the backwards direction has shown the condition is necessary, not that the chosen $\delta$ is sufficient — though for a chain of reversible inequalities, as here, both directions coincide and either can be written up.

3. A Harder Example: A Nonlinear Function

Linear functions make $\delta$ exactly proportional to $\varepsilon$. For nonlinear $f$, the factor relating $|x-a|$ to $|f(x)-L|$ is itself not constant, so we must first bound it on some neighbourhood of $a$, which is why a first restriction on $\delta$ (e.g. $\delta \le 1$) appears before the final choice.

Worked example. Prove that $\lim_{x\to 2} x^2 = 4$.

Scratch work. $|x^2-4| = |x-2|\,|x+2|$. The factor $|x+2|$ is not a constant, but if we first insist $\delta \le 1$, then $0<|x-2|<1$ gives $1<x<3$, hence $3<x+2<5$, so $|x+2|<5$. Then $|x^2-4| < 5|x-2|$, and we want this $<\varepsilon$, i.e. $|x-2|<\varepsilon/5$. So we need $\delta \le 1$ and $\delta \le \varepsilon/5$; take the minimum.

Proof. Let $\varepsilon>0$. Choose $\delta = \min\{1, \varepsilon/5\}$. Suppose $0<|x-2|<\delta$. Since $\delta\le 1$, we have $|x-2|<1$, so $1<x<3$ and thus $|x+2|<5$. Then

Since $\varepsilon>0$ was arbitrary, $\lim_{x\to2}x^2=4$. $\blacksquare$

The device "take $\delta$ to be the minimum of a bound that controls the troublesome factor and a bound that delivers the target $\varepsilon$" recurs throughout analysis; remember it, not just this one computation.

4. Uniqueness of the Limit

Theorem. If $\lim_{x\to a}f(x) = L_1$ and $\lim_{x\to a}f(x)=L_2$, then $L_1 = L_2$.

Proof. Suppose for contradiction $L_1 \ne L_2$. Let $\varepsilon = \dfrac{|L_1-L_2|}{2} > 0$. By the two limit hypotheses applied to this $\varepsilon$, there exist $\delta_1,\delta_2>0$ such that $0<|x-a|<\delta_1 \implies |f(x)-L_1|<\varepsilon$ and $0<|x-a|<\delta_2 \implies |f(x)-L_2|<\varepsilon$. Let $\delta=\min\{\delta_1,\delta_2\}$, and pick any $x$ with $0<|x-a|<\delta$ (such $x$ exists, e.g. $x = a+\delta/2$). Then by the triangle inequality,

a contradiction ($|L_1-L_2|<|L_1-L_2|$ is false). Hence $L_1=L_2$. $\blacksquare$

This is why the notation $\lim_{x\to a}f(x)=L$ is legitimate: the limit, when it exists, is a single well-defined number, and "the limit does not exist" is a meaningful alternative, not a failure of notation.

5. One-Sided Limits

Definition. $\lim_{x\to a^+} f(x) = L$ means: for every $\varepsilon>0$ there is $\delta>0$ such that $a < x < a+\delta \implies |f(x)-L|<\varepsilon$. Similarly $\lim_{x\to a^-}f(x)=L$ restricts to $a-\delta<x<a$.

Theorem. $\lim_{x\to a}f(x) = L$ if and only if both one-sided limits exist and $\lim_{x\to a^+}f(x) = \lim_{x\to a^-}f(x) = L$.

The proof is a direct unpacking of the definitions: the two-sided condition $0<|x-a|<\delta$ is exactly the union of $a-\delta<x<a$ and $a<x<a+\delta$, so a single $\delta$ working two-sidedly is equivalent to $\delta$'s working on each side separately (take the minimum of the two one-sided $\delta$'s, or re-use one $\delta$ for both). This theorem is the standard tool for showing a limit fails to exist at a jump: compute both one-sided limits and show they disagree.

Worked example. For $f(x) = \dfrac{|x|}{x}$ ($x\ne0$), $\lim_{x\to0^+}f(x) = \lim_{x\to0^+}1 = 1$ and $\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-1)=-1$. Since $1\ne-1$, $\lim_{x\to0}f(x)$ does not exist.

6. Limits at Infinity and Infinite Limits

Definition. $\lim_{x\to\infty}f(x)=L$ means: for every $\varepsilon>0$ there exists $M>0$ such that $x>M \implies |f(x)-L|<\varepsilon$. Here $M$ plays the role $\delta$ played before — it bounds how large $x$ must be, rather than how close to $a$.

Definition. $\lim_{x\to a}f(x) = \infty$ means: for every $K>0$ there exists $\delta>0$ such that $0<|x-a|<\delta \implies f(x) > K$. This is not a claim that the limit exists in the sense of Section 1 — $\infty$ is not a real number — it is a separate piece of notation recording a specific kind of divergence.

A common confusion. "The limit is $\infty$" and "the limit does not exist" are both true simultaneously for, e.g., $f(x)=1/x^2$ at $a=0$: no real number $L$ satisfies the Section 1 definition, yet the specific divergence is informative enough to deserve its own symbol and its own quantifier pattern.

7. Sequential Characterisation of the Limit

Theorem (sequential criterion). $\lim_{x\to a}f(x) = L$ if and only if for every sequence $(x_n)$ with $x_n \ne a$ for all $n$ and $x_n \to a$, we have $f(x_n) \to L$.

Proof ($\Rightarrow$). Suppose $\lim_{x\to a}f(x)=L$ and let $(x_n)\to a$, $x_n\ne a$. Given $\varepsilon>0$, choose $\delta>0$ from the hypothesis so that $0<|x-a|<\delta\implies|f(x)-L|<\varepsilon$. Since $x_n\to a$, there is $N$ such that $n>N \implies |x_n-a|<\delta$; also $x_n\ne a$ gives $0<|x_n-a|$. Hence $n>N \implies |f(x_n)-L|<\varepsilon$, i.e. $f(x_n)\to L$.

Proof ($\Leftarrow$, by contrapositive). Suppose $\lim_{x\to a}f(x) \ne L$ (or does not exist equal to $L$). Then there is some $\varepsilon_0>0$ such that no $\delta>0$ works: for every $\delta>0$ there is an $x$ with $0<|x-a|<\delta$ but $|f(x)-L|\ge\varepsilon_0$. Apply this with $\delta = 1/n$ for each $n\in\mathbb{N}$ to construct $x_n$ with $0<|x_n-a|<1/n$ and $|f(x_n)-L|\ge\varepsilon_0$. Then $x_n\to a$, $x_n\ne a$, but $f(x_n)\not\to L$. This is the contrapositive of the claim, so the claim holds. $\blacksquare$

This theorem is the standard tool for proving a limit does not exist without guessing an $\varepsilon_0$ directly: exhibit two sequences $x_n\to a$ and $y_n \to a$ with $f(x_n)\to L_1 \ne L_2 \leftarrow \lim f(y_n)$.

Worked example. $\lim_{x\to0}\sin(1/x)$ does not exist. Take $x_n = \dfrac{1}{2\pi n}\to 0$, giving $f(x_n)=\sin(2\pi n)=0$. Take $y_n = \dfrac{1}{\pi/2+2\pi n}\to0$, giving $f(y_n)=\sin(\pi/2+2\pi n)=1$. Since $0\ne1$, the sequential criterion shows the limit does not exist — note that the function is perfectly bounded, so this has nothing to do with blowing up, only with oscillation.

Common Exam Pitfalls

• Reversing the quantifier order. Writing "there exists $\delta$ such that for all $\varepsilon$..." instead of "for all $\varepsilon$ there exists $\delta$..." is a different (false, for non-constant $f$) statement, not a stylistic variant.
• Choosing $\delta$ before bounding the nonlinear factor. For $f(x)=x^2$ or similar, skipping the preliminary restriction $\delta\le1$ step leaves an unbounded factor like $|x+2|$ loose in the final inequality.
• Forgetting $x\ne a$. The punctured condition $0<|x-a|<\delta$ is essential; dropping it silently assumes $f(a)=L$, which need not hold and is irrelevant to the limit's existence.
• Treating $\lim_{x\to a}f(x)=\infty$ as "the limit exists". It is standard shorthand for a specific divergence pattern, not a value satisfying the Section 1 definition — say so explicitly if asked whether the limit "exists".
• Using one sequence to prove a limit exists. The sequential criterion requires every sequence $x_n\to a$; checking one or two convergent sequences can only ever disprove a limit (by exhibiting disagreement), never prove it exists.
• Confusing the roles of $\varepsilon$ and $\delta$ with $\delta$ "depending on $x$" instead of on $\varepsilon$. $\delta$ may depend on $a$ and $\varepsilon$ only; it must work uniformly for all $x$ in the punctured neighbourhood, not be re-chosen per $x$.

Summary

• Definition: $\lim_{x\to a}f(x)=L$ iff $\forall\varepsilon>0\,\exists\delta>0$ such that $0<|x-a|<\delta \implies |f(x)-L|<\varepsilon$.
• To find $\delta$: manipulate $|f(x)-L|$ algebraically into (constant)$\times|x-a|$, restricting $\delta\le1$ first if a nonlinear factor needs bounding, then take $\delta=\min\{\text{bound}, \varepsilon/\text{constant}\}$.
• Uniqueness: a limit, if it exists, is unique — proved by splitting $\varepsilon=|L_1-L_2|/2$ and the triangle inequality.
• One-sided limits $\lim_{x\to a^\pm}f(x)$ restrict $x$ to one side of $a$; the two-sided limit exists iff both one-sided limits exist and agree.
• Limits at infinity / infinite limits replace $\delta$ (or the target inequality) with $M$ or $K$, but keep the same $\forall$-$\exists$ logical shape.
• Sequential criterion: $\lim_{x\to a}f(x)=L$ iff $f(x_n)\to L$ for every sequence $x_n\to a$ with $x_n\ne a$ — the standard tool for proving non-existence via two sequences with different limits.