Continuity

Chapter 1.1 insisted that $\lim_{x \to a} f(x)$ and $f(a)$ are different questions. Continuity is the name for the situation where they happen to have the same answer — and that is the situation in which the whole of move 1, substitution, is legal.

$f$ is continuous at $a$ when

$$\lim_{x \to a} f(x) = f(a)$$

which is really three demands folded into one line, and it is worth unfolding them because an exam question usually breaks exactly one:

1. $f(a)$ exists — $a$ is in the domain.
2. $\lim_{x \to a} f(x)$ exists — both one-sided limits agree.
3. They are equal.

The three ways it breaks

Each failure has a name, and naming it is often the question.

Removable. The limit exists but the value does not match it — either $f(a)$ is undefined, or it is defined to be something else. The graph has a hole.

$$f(x) = \frac{x^2-1}{x-1} \quad\text{at } x = 1$$

The limit is $2$; $f(1)$ does not exist. Defining $f(1) = 2$ repairs it completely, which is why it is called removable.

Jump. The one-sided limits both exist and disagree, so condition 2 fails. No redefinition at the single point can help — the problem is on both sides of it, not at it.

Infinite. The function grows without bound, so condition 2 fails in a different way. $\tfrac{1}{x^2}$ at $0$, or $\tan x$ at $\tfrac{\pi}{2}$.

Making a function continuous on purpose

The standard exam question: a piecewise function with an unknown constant, and “find $k$ so that $f$ is continuous”. The method follows straight from the three conditions — set the one-sided limits equal to each other and to the value.

For

$$f(x) = \begin{cases} x^2 + k & x \le 2 \\ 3x - 1 & x > 2 \end{cases}$$

the left limit is $4 + k$, the right limit is $5$, and continuity demands they match: $k = 1$.

That verified line is the condition, written out at the joint. If the two sides do not agree there, no choice of $k$ makes them.

The Intermediate Value Theorem

If $f$ is continuous on the closed interval $[a, b]$, and $N$ is any value between $f(a)$ and $f(b)$, then $f(c) = N$ for at least one $c$ in $(a, b)$.

Informally: a continuous function cannot get from one value to another without passing through everything in between. It is obvious once stated and it is the tool behind every “show that this equation has a solution” question.

To show $x^3 + x - 1 = 0$ has a root between $0$ and $1$: the function is a polynomial, so continuous; at $0$ it is $-1$; at $1$ it is $1$. Zero lies between them, so some $c$ in $(0,1)$ gives exactly zero.

Those two evaluations are the entire proof. The theorem does the rest.