Continuity

The three-part continuity check

A function $f$ is continuous at $x = a$ when all three of these hold:

$f(a)$ is defined.
$\lim_{x \to a} f(x)$ exists, meaning the left-hand and right-hand limits are equal.
That limit equals $f(a)$ .

Each kind of discontinuity is a different one of these failing:

Kinds of discontinuity and the condition each one breaks

Type	What fails	Example
Jump	Left and right limits exist but differ	$\lfloor x \rfloor$ at any integer
Removable	The limit exists but $f(a)$ is missing or different	$\dfrac{x^2 - 1}{x - 1}$ at $x = 1$
Infinite	The function grows without bound	$\dfrac{1}{x}$ at $x = 0$

A jump at x = 2, a removable gap at x = 5 (the curve's point is missing and f(5) sits elsewhere) and a corner at x = 7.5. The corner is continuous; it fails only differentiability.

Worked example: choosing a constant for continuity

Find $k$ so that $f(x) = \begin{cases} kx + 1, & x \le 2 \\ 3x - 1, & x > 2 \end{cases}$ is continuous at $x = 2$ .

Solution: $f(2) = 2k + 1$ , which is also the left-hand limit. The right-hand limit is $3(2) - 1 = 5$ . For continuity they must match:

2k + 1 = 5 \;\Rightarrow\; k = 2

A function can be continuous but not differentiable

A function is continuous at a point if there's no break or jump there, but it's only differentiable at that point if the slope approaching from the left matches the slope approaching from the right. The left-hand derivative is:

f'(a^-) = \lim_{h \to 0^-} \dfrac{f(a+h) - f(a)}{h}

Worked example

Find the left-hand derivative of $f(x) = |x|$ at $x = 0$ .

Solution: Just to the left of 0, $x < 0$ , so $f(x) = |x| = -x$ . Differentiating this piece:

f'(x) = -1 \quad \text{for } x < 0

So the left-hand derivative at $x=0$ is -1. (The right-hand derivative there is $+1$ , since $f(x)=x$ for $x>0$ , the two don't match, which is exactly why $|x|$ is continuous at 0 but has a sharp corner there instead of a smooth tangent.)

A function can only be continuous where it's defined

Before asking whether a function is continuous at a point, it has to actually be defined there, so finding a function's domain is usually the first step.

Find the domain of $f(x) = \sqrt{\cos x}$ over one full period, $[0, 2\pi]$ .

Solution: A square root is only defined for a non-negative input, so $f(x)$ requires $\cos x \ge 0$ . Over $[0, 2\pi]$ , cosine is non-negative on the first and last quarters of the circle:

\left[0, \dfrac{\pi}{2}\right] \cup \left[\dfrac{3\pi}{2}, 2\pi\right]

Common mistakes

Checking only one side. A piecewise function needs both one-sided limits at the joining point.
Assuming continuous means differentiable. $|x|$ is continuous at 0 but has a corner there. Differentiable does imply continuous, though.
Calling a point outside the domain a discontinuity. $\sqrt{x}$ isn't discontinuous at $x = -1$ . It simply isn't defined there.