Continuity

By the end you should be able to: State the three conditions for continuity at a point, classify discontinuities as removable or non-removable, and determine values that make a piecewise function continuous.

Informally: a function is continuous if you can draw its graph without lifting your pen. That intuition is right, and the formal version says exactly the same thing with no ambiguity.

The definition

$f$ is continuous at $x = a$ when

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

That single equation quietly demands three things, and an exam question will expect all three named:

$f(a)$ is defined — $a$ is in the domain
$\lim_{x\to a} f(x)$ exists — both one-sided limits agree
The two are equal

Fail any one and the function is discontinuous at $a$ .

Mapped onto the previous topic

The three ways to fail correspond to three pictures:

Failure	Picture	Example at $x=0$
$f(a)$ undefined	hole	$f(x) = \dfrac{\sin x}{x}$
Limit does not exist	jump	$f(x) = \dfrac{\\|x\\|}{x}$
Both exist but differ	hole with a stray point	$f(x)=x+1$ for $x\neq 0$ , $f(0)=5$

Removable and non-removable

A discontinuity is removable when the limit exists — you can patch the function at that single point and continuity is restored.

f(x) = \frac{x^2-4}{x-2}

is undefined at $x = 2$ , but $\lim_{x\to 2} f(x) = 4$ . Define $f(2) = 4$ and the function becomes continuous. The hole is filled.

A jump discontinuity is not removable. With $|x|/x$ the one-sided limits are $-1$ and $1$ ; no single value at $0$ can bridge them. You cannot patch a gap of width 2 with one point.

Continuity on an interval

$f$ is continuous on $[a,b]$ if it is continuous at every interior point, and one-sidedly continuous at the ends:

\lim_{x\to a^+} f(x) = f(a), \qquad \lim_{x\to b^-} f(x) = f(b)

Only one side is available at an endpoint, so only one side is required.

Which functions are continuous

On their domains: polynomials, $\sin$ , $\cos$ , $e^x$ , and rational functions wherever the denominator is non-zero. Sums, products, quotients and compositions of continuous functions are continuous.

This is why substitution works for limits of polynomials. It is not a separate rule — it is continuity, used.

The usual exam question

Find $k$ so that this piecewise function is continuous. The method is always the same: force the one-sided limits and the assigned value to agree, then solve for $k$ .