Limit Theorems and the Continuity of Functions

The $\varepsilon$-$\delta$ definition of the previous chapter is precise but painful to use directly on every function you meet. This chapter proves a small toolbox of limit laws — for sums, products, quotients and compositions — each derived once and for all from the definition, so that afterwards you compute limits of complicated expressions algebraically instead of re-running an $\varepsilon$-$\delta$ argument every time. The second half of the chapter uses these laws to define continuity rigorously and to prove the two theorems that make continuous functions on closed intervals so useful: the Intermediate Value Theorem and the Extreme Value Theorem.

Have to hand: the $\varepsilon$-$\delta$ definition of a limit, the triangle inequality, and the notion of a bounded set. Everything here is built on Chapter 1 and nothing is assumed beyond it.

What continuity is not. Continuity is often introduced as "you can draw the graph without lifting your pen." That is a useful cartoon but not a definition — it cannot be checked for a function you cannot draw, and it says nothing about why the theorems of this chapter hold. The precise definition in Section 4 is what every proof below actually uses.

1. The Limit Laws

Theorem (Sum Rule). If $\lim_{x\to a}f(x)=L_1$ and $\lim_{x\to a}g(x)=L_2$, then $\lim_{x\to a}[f(x)+g(x)] = L_1+L_2$.

Proof. Let $\varepsilon>0$. Apply the hypothesis on $f$ with tolerance $\varepsilon/2$: there is $\delta_1>0$ such that $0<|x-a|<\delta_1 \implies |f(x)-L_1|<\varepsilon/2$. Apply the hypothesis on $g$ likewise: there is $\delta_2>0$ such that $0<|x-a|<\delta_2 \implies |g(x)-L_2|<\varepsilon/2$. Let $\delta=\min\{\delta_1,\delta_2\}$. If $0<|x-a|<\delta$, both conditions hold simultaneously, and by the triangle inequality,

Since $\varepsilon>0$ was arbitrary, the sum rule holds. $\blacksquare$

Theorem (Product Rule). If $\lim_{x\to a}f(x)=L_1$ and $\lim_{x\to a}g(x)=L_2$, then $\lim_{x\to a}[f(x)g(x)] = L_1L_2$.

Proof sketch. Write $f(x)g(x)-L_1L_2 = f(x)[g(x)-L_2] + L_2[f(x)-L_1]$. By applying the definition of $\lim_{x\to a}f(x)=L_1$ with $\varepsilon=1$ we obtain $\delta_0>0$ with $0<|x-a|<\delta_0\implies |f(x)-L_1|<1$, hence $|f(x)|<|L_1|+1=:M$ on that punctured neighbourhood. Given $\varepsilon$, make $|g(x)-L_2|<\varepsilon/(2M)$ and $|f(x)-L_1|<\varepsilon/(2(|L_2|+1))$ using the two hypotheses, then take the minimum of the resulting $\delta$'s (together with the $\delta$ that gave the bound $M$). The triangle inequality closes the estimate exactly as in the sum rule. $\blacksquare$

Theorem (Quotient Rule). If $\lim_{x\to a}f(x)=L_1$, $\lim_{x\to a}g(x)=L_2\ne0$, then $\lim_{x\to a}\dfrac{f(x)}{g(x)} = \dfrac{L_1}{L_2}$.

Key step. Since $L_2\ne0$, taking $\varepsilon = |L_2|/2$ in the hypothesis on $g$ gives a $\delta_0$ such that $0<|x-a|<\delta_0 \implies |g(x)|>|L_2|/2$, so $g(x)$ is bounded away from zero near $a$ and $1/g(x)$ is well-defined and bounded there. The rest follows the product rule pattern applied to $f(x)\cdot\frac{1}{g(x)}$, after first proving $\lim_{x\to a}\frac{1}{g(x)}=\frac{1}{L_2}$ directly from $|1/g(x)-1/L_2| = |g(x)-L_2|/(|g(x)||L_2|)$.

Worked example. Compute $\lim_{x\to2}\dfrac{3x^2-5}{x+1}$. Since $\lim_{x\to2}(3x^2-5) = 3(4)-5=7$ (by the product and sum rules applied to the basic limit $\lim_{x\to2}x=2$) and $\lim_{x\to2}(x+1)=3\ne0$, the quotient rule gives $\lim_{x\to2}\dfrac{3x^2-5}{x+1} = \dfrac{7}{3}$.

Why these proofs matter even though you never re-derive them. Once proved, the sum, product and quotient rules let you compute $\lim_{x\to a}(\text{any rational function})$ by direct substitution, provided the denominator does not vanish at $a$ — that proviso is exactly where the quotient rule's hypothesis $L_2\ne0$ bites.

2. The Squeeze Theorem

Theorem (Squeeze/Sandwich Theorem). Suppose $g(x) \le f(x) \le h(x)$ for all $x$ in some punctured neighbourhood of $a$, and $\lim_{x\to a}g(x) = \lim_{x\to a}h(x) = L$. Then $\lim_{x\to a}f(x) = L$.

Proof. Let $\varepsilon>0$. Choose $\delta_1,\delta_2>0$ from the hypotheses on $g,h$ so that $0<|x-a|<\delta_1 \implies L-\varepsilon<g(x)<L+\varepsilon$ and $0<|x-a|<\delta_2 \implies L-\varepsilon<h(x)<L+\varepsilon$. Let $\delta=\min\{\delta_1,\delta_2,\delta_3\}$ where $\delta_3$ is the radius of the neighbourhood on which $g\le f\le h$ holds. For $0<|x-a|<\delta$,

so $|f(x)-L|<\varepsilon$. $\blacksquare$

Worked example. Prove $\lim_{x\to0}x^2\sin(1/x) = 0$. Since $-1\le\sin(1/x)\le1$ for $x\ne0$, multiplying by $x^2\ge0$ gives $-x^2 \le x^2\sin(1/x) \le x^2$. Both bounding functions tend to $0$ as $x\to0$ (direct from the product rule), so the Squeeze Theorem gives $\lim_{x\to0}x^2\sin(1/x)=0$ — even though $\sin(1/x)$ itself has no limit at $0$ (Chapter 1, Section 7), the shrinking factor $x^2$ forces the product to converge anyway.

3. Basic Limits by Direct Substitution

Combining the limit laws with the trivial limits $\lim_{x\to a}c = c$ and $\lim_{x\to a}x = a$ (both immediate from the $\varepsilon$-$\delta$ definition) gives, by induction on the number of operations:

Corollary. If $p(x)$ is a polynomial, $\lim_{x\to a}p(x) = p(a)$ for every $a$. If $r(x)=p(x)/q(x)$ is a rational function and $q(a)\ne0$, $\lim_{x\to a}r(x) = r(a)$.

This single corollary is why direct substitution works for "nice" functions — it is not a separate axiom, it is a consequence of the sum, product and quotient rules applied repeatedly.

4. The Definition of Continuity

Definition. $f$ is continuous at $a$ if (i) $f(a)$ is defined, (ii) $\lim_{x\to a}f(x)$ exists, and (iii) $\lim_{x\to a}f(x) = f(a)$. Equivalently, in a single quantified statement:

Notice the condition is now $|x-a|<\delta$, not punctured — $x=a$ is allowed and trivially satisfies the conclusion since $|f(a)-f(a)|=0$. $f$ is continuous on a set $S$ if it is continuous at every point of $S$; continuous (unqualified) usually means continuous at every point of its domain.

The three conditions, and where each is used to rule out a discontinuity. A removable discontinuity fails exactly condition (i) or has (ii) true but (iii) false (limit exists, but $f(a)$ is undefined or wrong). A jump discontinuity fails (ii): the one-sided limits exist but disagree. An infinite/essential discontinuity fails (ii) in a stronger way: no finite limit exists at all.

Theorem (continuity and the limit laws). If $f,g$ are continuous at $a$, so are $f+g$, $fg$, and (if $g(a)\ne0$) $f/g$. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f\circ g$ is continuous at $a$.

Proof (composition case). Let $\varepsilon>0$. Since $f$ is continuous at $g(a)$, there is $\eta>0$ such that $|y-g(a)|<\eta \implies |f(y)-f(g(a))|<\varepsilon$. Since $g$ is continuous at $a$, applying this $\eta$ as the tolerance gives $\delta>0$ such that $|x-a|<\delta \implies |g(x)-g(a)|<\eta$. Chaining the two implications: $|x-a|<\delta \implies |g(x)-g(a)|<\eta \implies |f(g(x))-f(g(a))|<\varepsilon$, which is exactly continuity of $f\circ g$ at $a$. $\blacksquare$

The other three parts follow immediately from the corresponding limit laws of Section 1, using $\lim_{x\to a}f(x)=f(a)$ and $\lim_{x\to a}g(x)=g(a)$ in place of general limit values $L_1,L_2$.

Corollary. Polynomials are continuous everywhere; rational functions are continuous on their domain (wherever the denominator is nonzero); compositions of continuous functions are continuous.

5. One-Sided Continuity and Classifying Discontinuities

Definition. $f$ is right-continuous at $a$ if $\lim_{x\to a^+}f(x)=f(a)$, and left-continuous if $\lim_{x\to a^-}f(x)=f(a)$. By Chapter 1, Section 5, $f$ is continuous at $a$ iff it is both left- and right-continuous there.

Worked example. Classify the discontinuity of $f(x) = \dfrac{x^2-1}{x-1}$ at $x=1$ (with $f(1)$ left undefined), versus $g(x)=\lfloor x\rfloor$ at $x=1$. For $f$: $\lim_{x\to1}f(x)=\lim_{x\to1}(x+1)=2$ exists, but $f(1)$ is undefined — a removable discontinuity (define $f(1)=2$ and $f$ becomes continuous there). For $g$: $\lim_{x\to1^-}g(x)=0$ and $\lim_{x\to1^+}g(x)=1$ disagree — a jump discontinuity, which no redefinition of $g(1)$ can repair, since the two one-sided limits themselves conflict.

6. The Intermediate Value Theorem

Theorem (IVT). Let $f$ be continuous on the closed interval $[a,b]$, and let $k$ be any number strictly between $f(a)$ and $f(b)$. Then there exists $c\in(a,b)$ with $f(c)=k$.

Proof idea (via the supremum). Assume $f(a)<k<f(b)$ (the other case is symmetric). Let $S=\{x\in[a,b]: f(x)<k\}$. $S$ is nonempty ($a\in S$) and bounded above by $b$, so by the completeness of $\mathbb{R}$, $c=\sup S$ exists. Continuity of $f$ at $c$, together with the definition of supremum, rules out both $f(c)<k$ (which would put points slightly to the right of $c$ still in $S$, contradicting $c=\sup S$) and $f(c)>k$ (which would exclude points slightly to the left of $c$ from being a supremum of $S$), leaving $f(c)=k$ as the only possibility. The full epsilon-delta details of this argument belong to a course in real analysis; what matters here is which hypotheses are load-bearing.

Every hypothesis of IVT is necessary. Continuity cannot be dropped: let $f(x)=-1$ for $x\in[-1,0)$ and $f(x)=1$ for $x\in[0,1]$. This $f$ is discontinuous at $x=0$ (jumping from $-1$ to $1$), and indeed $0$ lies strictly between $f(-1)=-1$ and $f(1)=1$ yet $f$ never equals $0$ anywhere on $[-1,1]$ — IVT's conclusion genuinely fails without continuity. (Note $\text{sign}(x)$ itself is not a valid counterexample here, since $\text{sign}(0)=0$ and the conclusion $f(c)=0$ does hold, at $c=0$.) The interval must be an interval (connected): continuity on a union of two disjoint intervals can skip values between them. And $[a,b]$ cannot be replaced by an open interval without additional care about endpoint behaviour, since $k$ is pinned between the endpoint values $f(a),f(b)$.

Worked example. Show $x^3 - 3x + 1 = 0$ has a root in $(1,2)$. Let $f(x)=x^3-3x+1$, continuous everywhere (a polynomial). $f(1) = 1-3+1=-1<0$ and $f(2)=8-6+1=3>0$. Since $f$ is continuous on $[1,2]$ and $0$ lies strictly between $f(1)=-1$ and $f(2)=3$, IVT guarantees some $c\in(1,2)$ with $f(c)=0$.

7. The Extreme Value Theorem and Boundedness

Theorem (Boundedness Theorem). If $f$ is continuous on the closed, bounded interval $[a,b]$, then $f$ is bounded on $[a,b]$.

Theorem (Extreme Value Theorem, EVT). If $f$ is continuous on $[a,b]$, then $f$ attains both a maximum and a minimum value on $[a,b]$: there exist $c,d\in[a,b]$ with $f(c)\ge f(x)\ge f(d)$ for all $x\in[a,b]$.

Both theorems rest crucially on $[a,b]$ being closed and bounded (compact). Each hypothesis is necessary: $f(x)=1/x$ is continuous but unbounded on the bounded-but-not-closed interval $(0,1]$; $f(x)=x$ is continuous but unbounded on the closed-but-not-bounded interval $[0,\infty)$; and $f(x)=x$ on the open interval $(0,1)$ never attains a maximum, approaching but never reaching $1$.

IVT, EVT and boundedness are existence theorems, not construction recipes. They guarantee a root, or a maximum, exists; they do not tell you where. Finding the actual location is the job of the derivative tools in the chapters that follow.

Common Exam Pitfalls

• Using the quotient rule without checking $L_2\ne0$. If the denominator's limit is $0$, the quotient rule simply does not apply; the limit may still exist (via cancellation) or be infinite — it must be investigated separately.
• Treating "the graph has no gaps" as the definition. It is an intuition for continuity, not the testable condition; always check the three numbered conditions explicitly when asked to justify continuity.
• Forgetting that continuity is a two-sided, non-punctured condition. Unlike the limit, $f(a)$ is squarely part of the continuity requirement — a removable discontinuity is a discontinuity precisely because the (existing) limit and the (defined or not) value disagree.
• Misapplying IVT when $f$ is not continuous on the whole interval. A single internal discontinuity, even "invisible" in a rough sketch, can break the guarantee entirely.
• Assuming EVT or the Boundedness Theorem applies on an open or unbounded interval. Compactness ($[a,b]$ closed and bounded) is not a technicality; EVT provably fails without it, as the counterexamples in Section 7 show. (IVT only needs continuity on the whole interval between the two points used.)
• Confusing a root guaranteed by IVT with a unique root. IVT guarantees existence of at least one $c$ with $f(c)=k$; it says nothing about uniqueness, which requires additional monotonicity information.

Summary

• Limit laws: sum, product and quotient rules reduce limits of combined functions to limits of the pieces, each proved once from the $\varepsilon$-$\delta$ definition via $\varepsilon/2$-splitting and the triangle inequality; the quotient rule requires the denominator's limit to be nonzero.
• Squeeze theorem: if $g\le f\le h$ near $a$ and $g,h\to L$, then $f\to L$ — useful exactly when $f$ itself resists direct estimation.
• Direct substitution for polynomials and (where the denominator is nonzero) rational functions is a consequence of the limit laws, not a separate rule.
• Continuity at $a$: $f(a)$ defined, $\lim_{x\to a}f(x)$ exists, and the two are equal — equivalently $\forall\varepsilon\exists\delta: |x-a|<\delta\implies|f(x)-f(a)|<\varepsilon$, with no puncture.
• Discontinuity types: removable (limit exists, value missing or wrong), jump (one-sided limits disagree), infinite/essential (no finite limit).
• Sums, products, quotients and compositions of continuous functions are continuous, hence polynomials, rational functions (off their poles) and compositions thereof are continuous.
• IVT: continuity on $[a,b]$ guarantees every value between $f(a)$ and $f(b)$ is attained somewhere in $(a,b)$ — an existence, not a location, result.
• EVT/Boundedness: continuity on a closed, bounded interval guarantees $f$ is bounded and attains both a maximum and a minimum; compactness of the interval is essential to both.