Continuity of Functions

Two chapters of limit technique now converge on a single, precise idea: what does it mean to say a function has "no breaks" in its graph? The everyday sense — "you could draw it without lifting your pen" — is a useful picture, but it is not a definition that can be tested algebraically. This chapter supplies that definition, built entirely from the limit vocabulary already in hand, and uses it to classify exactly how a function can fail to be continuous.

Continuity matters beyond its own sake: the derivative, introduced in the next chapter, cannot exist at a point where the function is not even continuous, so continuity is the gatekeeping condition for everything that follows in this block. Have the one-sided limit, hole and asymptote vocabulary from the previous two chapters ready.

1. The Three-Part Definition of Continuity at a Point

Definition. A function $f$ is continuous at $x=a$ if all three of the following hold:

1. $f(a)$ is defined,
2. $\displaystyle\lim_{x\to a} f(x)$ exists,
3. $\displaystyle\lim_{x\to a} f(x) = f(a)$.

Each condition rules out a different kind of failure. Condition 1 rules out a point where the function simply has no value (a hole, or a point outside the domain). Condition 2 rules out one-sided limits that disagree, or a limit that is infinite. Condition 3 is the one most often overlooked: a function can satisfy the first two conditions and still fail to be continuous, if its actual value at $a$ does not match the value its neighbours are converging towards.

Worked example. Is $f(x) = \begin{cases} x^2 & x\neq2 \\ 10 & x=2 \end{cases}$ continuous at $x=2$?

Condition 1: $f(2)=10$ is defined. Condition 2: $\displaystyle\lim_{x\to2} x^2 = 4$ exists (the piecewise redefinition at the single point $x=2$ does not affect the limit, which only cares about nearby values). Condition 3: is $\displaystyle\lim_{x\to2}f(x) = f(2)$? Here $4 \neq 10$, so condition 3 fails, and $f$ is not continuous at $x=2$ — despite being defined there and having a perfectly well-behaved limit.

2. Continuity on an Interval

Definition. $f$ is continuous on an interval $I$ if it is continuous at every point of $I$. At an endpoint of a closed interval, only the relevant one-sided limit is required to match the function value.

A function continuous on its entire domain is simply called continuous. Polynomials are continuous everywhere, since direct substitution always succeeds (Chapter 1, Section 4). Rational functions are continuous everywhere except at the zeros of their denominator — precisely the points classified as holes or vertical asymptotes in the previous chapter.

3. Classifying Discontinuities

A function fails to be continuous at $a$ in one of three distinguishable ways, each corresponding to a different condition in the definition failing.

Removable discontinuity. $\displaystyle\lim_{x\to a}f(x)$ exists, but either $f(a)$ is undefined or $f(a)$ differs from the limit. This is exactly a hole: the discontinuity could be "fixed" by defining or redefining $f(a)$ to equal the limit.

Jump discontinuity. Both one-sided limits exist but disagree: $\displaystyle\lim_{x\to a^-}f(x) \neq \lim_{x\to a^+}f(x)$. The two-sided limit does not exist, and no single redefinition of $f(a)$ can repair it, since the function genuinely jumps between two different levels.

Infinite discontinuity. At least one one-sided limit is $+\infty$ or $-\infty$ — exactly a vertical asymptote. No finite redefinition of $f(a)$ could ever make this continuous.

Worked example. Classify the discontinuity of each function at the stated point:

(a) $f(x) = \dfrac{x^2-9}{x-3}$ at $x=3$: the limit exists ($=6$) but $f(3)$ is undefined — removable.

(b) $g(x) = \begin{cases} x & x<1 \\ x+2 & x\geq1 \end{cases}$ at $x=1$: $\displaystyle\lim_{x\to1^-}g(x)=1$, $\displaystyle\lim_{x\to1^+}g(x)=3$, these disagree — jump.

(c) $h(x) = \dfrac1{(x-2)^2}$ at $x=2$: both one-sided limits are $+\infty$ — infinite.

4. Why Jump Discontinuities Cannot Be Removed

A removable discontinuity earns its name because a single new definition of $f(a)$ — setting it equal to the limit — makes the function continuous there. A jump discontinuity admits no such fix: whatever value is assigned to $f(a)$, it can match at most one of the two one-sided limits, never both simultaneously (unless, trivially, they already agreed, in which case there was no jump to begin with). This is the formal reason piecewise functions built from mismatched branches can never be patched into continuity at the seam.

5. Continuity of Combinations of Functions

Because continuity is defined entirely through limits, and limits respect sums, products and quotients (Chapter 1, Section 4), continuity inherits the same structure:

Theorem. If $f$ and $g$ are both continuous at $a$, then so are $f+g$, $f-g$, $fg$, and $\dfrac{f}{g}$ (provided $g(a)\neq0$). If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then the composite $f\circ g$ is continuous at $a$.

This theorem is what justifies, without further work, that an expression such as $\sin(x^2+1)$ or $\dfrac{x+3}{x^2+5}$ is continuous everywhere it is defined: it is built entirely from continuous pieces (polynomials, and — anticipating Chapter 8 — standard functions like $\sin$) combined by operations that preserve continuity.

6. The Intermediate Value Theorem

Continuity on an interval has a powerful geometric consequence: a continuous function cannot skip over any value between two of its outputs without actually attaining it somewhere in between.

Intermediate Value Theorem (IVT). If $f$ is continuous on $[a,b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in [a,b]$ with $f(c) = k$.

The theorem is a direct consequence of there being no gaps to jump across: if $f$ never equalled $k$, the graph would have to leap from one side of the line $y=k$ to the other without crossing it, which is exactly the behaviour continuity forbids. The IVT is not constructive — it proves such a $c$ exists without saying how to find it — but it is the theoretical justification behind every "root between two values" argument and underlies numerical root-finding methods such as the bisection method.

Worked example. Show that $f(x) = x^3 - x - 3$ has a root between $x=1$ and $x=2$.

$f$ is a polynomial, hence continuous everywhere, in particular on $[1,2]$. Evaluate the endpoints:

Since $f(1) = -3 < 0 < 3 = f(2)$, and $0$ lies between $f(1)$ and $f(2)$, the IVT guarantees some $c\in(1,2)$ with $f(c)=0$ — a root exists in that interval, even though its exact value is not produced by this argument.

7. Continuity and the Domain of Validity for Limit Shortcuts

Key distinction. "$f$ is continuous at $a$" and "$\displaystyle\lim_{x\to a} f(x) = f(a)$" are the same statement, phrased two ways. This is precisely why, for a continuous function, a limit can always be evaluated by direct substitution — and precisely why direct substitution fails at a discontinuity, demanding the factor-and-cancel or rationalising techniques of Chapter 1.

This reframes the entire project of Chapter 1: resolving a $\frac00$ limit was always secretly a question about the behaviour of a function at a point of (removable) discontinuity.

Common Exam Pitfalls

• Checking only that $f(a)$ is defined and declaring continuity. All three conditions must hold; a function can be defined at $a$, have a limit at $a$, and still fail to be continuous if the two values disagree (Section 1's worked example).
• Assuming every discontinuity is a hole. Jump and infinite discontinuities are common and fundamentally different: neither can be patched by redefining a single function value.
• Trying to "fix" a jump discontinuity by averaging the one-sided limits. No single value of $f(a)$ can match two different limits simultaneously; a jump discontinuity is permanent, not removable.
• Misapplying the IVT when the function is not continuous on the whole interval. If $f$ has even one vertical asymptote inside $[a,b]$, the theorem's hypothesis fails and its conclusion cannot be relied upon.
• Treating the IVT as constructive. It proves a root (or intermediate value) exists; it does not locate it. Do not present "therefore $c=1.5$" as a conclusion the theorem itself supplies.
• Forgetting that piecewise functions must be checked at the seam, not elsewhere. Each branch is typically continuous on its own open interval by construction; the only point requiring real work is where the branches meet.

Summary

• $f$ is continuous at $a$ iff $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the two are equal.
• A removable discontinuity (hole) has an existing limit that disagrees with, or lacks, a function value — fixable by redefinition.
• A jump discontinuity has disagreeing one-sided limits — never fixable by any single redefinition.
• An infinite discontinuity is a vertical asymptote — never fixable, since no finite value could match an infinite limit.
• Sums, differences, products, quotients (nonzero denominator) and compositions of continuous functions are continuous; this justifies continuity of complex expressions without rechecking every point.
• The Intermediate Value Theorem: a function continuous on $[a,b]$ attains every value between $f(a)$ and $f(b)$ somewhere in $[a,b]$ — used to prove existence of roots without finding them exactly.
• Continuity at $a$ is identical to the statement that the limit at $a$ can be found by direct substitution; every $\frac00$ limit from Chapter 1 was a question about removable discontinuity.