Continuity

01The three-part definition

A function $f$ is continuous at $x = a$ when three things hold, and each can fail independently:

1. $f(a)$ is defined — the point exists.
2. $\lim\limits_{x \to a} f(x)$ exists — both one-sided limits agree.
3. $\lim\limits_{x \to a} f(x) = f(a)$ — the limit equals the actual value.

The intuition is exactly what you'd guess: you can draw the graph through $x = a$ without lifting your pen. Condition 3 is the one students forget to check — a function can be defined at $a$ and have a limit at $a$, yet be discontinuous because the two don't match. Checking continuity is literally checking these three boxes in order, and exam problems are graded that way.

02The three ways continuity fails

Each failed condition produces its own species of discontinuity:

• Removable (hole) — the limit exists but $f(a)$ is undefined or mismatched. Example: $\dfrac{x^2-9}{x-3}$ at $x=3$. One point is wrong; everything else is fine.
• Jump — both one-sided limits exist but disagree (the hero graph above). Typical of piecewise functions like $\dfrac{|x|}{x}$.
• Infinite — the function blows up to $\pm\infty$; a vertical asymptote, like $\dfrac{1}{x-2}$ at $x=2$.

The names matter: "removable" means you could repair the function by redefining a single point. Jumps and asymptotes can't be patched — no single value fixes them.

03Where the familiar functions are continuous

Good news: almost everything you work with is continuous on its domain. Polynomials are continuous everywhere. Rational functions are continuous except where the denominator is zero. Root, exponential, logarithmic, and trig functions are continuous at every point of their domains. Better still, continuity survives arithmetic: sums, differences, products, quotients (where defined), and compositions of continuous functions are continuous.

The practical consequence: to find the discontinuities of a typical function, just find where its domain breaks — denominators hitting zero, even roots of negatives, logs of non-positives. That is why the domain drills in Module 01 keep paying rent.

04Repairing a removable discontinuity

If $\lim\limits_{x\to a} f(x) = L$ exists but $f(a)$ is missing or wrong, define (or redefine) $f(a) = L$ and the function becomes continuous there. Example: $f(x) = \dfrac{x^2 - 4}{x - 2}$ is undefined at $2$, but the limit there is $4$, so the repaired function

$$g(x) = \begin{cases} \dfrac{x^2-4}{x-2} & x \ne 2 \\[4pt] 4 & x = 2 \end{cases}$$

is continuous everywhere. This "fill in the hole with the limit" move is a standard exam question and a preview of how mathematicians extend functions.

05Making piecewise functions continuous

The classic problem: find the constant $k$ that glues two pieces together. The recipe is always the same — set the two one-sided limits at the seam equal to each other (and to the function's value there), then solve for $k$. For

$$f(x) = \begin{cases} kx + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}$$

continuity at $x = 2$ requires $\lim\limits_{x\to 2^-} f(x) = \lim\limits_{x\to 2^+} f(x)$, i.e. $2k + 1 = 4$, so $k = \tfrac{3}{2}$. Two unknown constants means two seams (or one seam plus another condition) and a small system of equations — same idea, more algebra.

06Continuity on intervals

A function is continuous on an open interval $(a,b)$ if it's continuous at every point inside. On a closed interval $[a,b]$ we only ask for one-sided continuity at the endpoints: right-continuity at $a$ ($\lim_{x\to a^+} f(x) = f(a)$) and left-continuity at $b$. That's why $\sqrt{x}$ counts as continuous on $[0, \infty)$ even though it has no left side at $0$. Closed-interval continuity is the precise hypothesis that the big theorems below demand — and the Extreme Value Theorem in Module 08 will demand it again.

07The Intermediate Value Theorem

Here's the payoff for all this care. IVT: if $f$ is continuous on $[a,b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a,b)$ with $f(c) = N$. A continuous function can't skip values — to get from one height to another, it must pass through everything in between.

The killer application is proving roots exist. Take $f(x) = x^3 + x - 1$. Then $f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $f$ is continuous (it's a polynomial) and $0$ lies between $-1$ and $1$, IVT guarantees some $c$ in $(0,1)$ with $f(c) = 0$. You've proven a root exists without finding it — which is exactly what the theorem is for. Note what IVT does not say: it doesn't find $c$, and it doesn't say $c$ is unique.

08Why continuity matters for what's next

Continuity is the admission ticket for nearly every major theorem ahead. Differentiability (Module 04) is a stronger condition: every differentiable function is continuous, but not conversely — $|x|$ is continuous at $0$ yet has no derivative there. The Extreme Value Theorem needs continuity on a closed interval to promise a max and min. The Mean Value Theorem needs it. The definite integral (Module 10) is guaranteed to exist for continuous functions. When a theorem fails in a homework problem, the first thing to check is whether a continuity hypothesis was quietly violated.