Continuity and discontinuity
Continuity and Discontinuity
Continuity and Discontinuity is a topic in Limits and Continuity in the California Calculus Standards. It is aligned to Standard 2.0, which requires students to demonstrate knowledge of both the formal definition and the graphical interpretation of continuity of a function.
A function is continuous at \(x=a\) when \(\lim_{x\to a} f(x)=f(a)\), so its graph has no break there. A discontinuity is where this fails — a removable hole, a jump, or an infinite (asymptote) discontinuity.
Theory
A function is continuous at \(x=a\) when its graph has no break there — you can draw through the point without lifting your pen. This page gives the three-part definition, shows how to read continuity from a graph, and classifies each discontinuity as removable, jump, or infinite.
A function \(f\) is continuous at \(a\) when all three conditions hold:
- \(f(a)\) is defined;
- \(\lim_{x\to a} f(x)\) exists;
- \(\lim_{x\to a} f(x)=f(a)\) — the limit equals the function value.
If any condition fails, \(f\) is discontinuous at \(a\). There are three common types:
- Removable (a hole): the limit exists but \(f(a)\) is missing or different.
- Jump: the one-sided limits exist but disagree.
- Infinite: the function blows up at a vertical asymptote.
Continuity at a point in one line:
For a rational function \(\dfrac{p(x)}{q(x)}\), the discontinuities are exactly the solutions of
How to test continuity at \(x=a\)
- Evaluate \(f(a)\). Is it defined?
- Find \(\lim_{x\to a} f(x)\) (check both sides). Does it exist?
- Compare. If the limit equals \(f(a)\), \(f\) is continuous; otherwise classify the break as removable, jump, or infinite.
Check all three parts of the definition.
| \(f(3)\) | \(=\) | \(3^2-1=8\ \text{(defined)}\) |
| \(\lim_{x\to 3}f(x)\) | \(=\) | \(8\ \text{(exists)}\) |
| \(f(3)\) | \(=\) | \(\lim_{x\to 3}f(x)\) |
All three hold, so \(f\) is continuous at \(x=3\).
Factor: \(\dfrac{(x-3)(x+3)}{x-3}=x+3\) for \(x\neq 3\). The limit is \(6\), but \(f(3)\) is undefined — a single missing point.
This is a removable discontinuity (a hole at \((3,6)\)).
A rational function breaks where its denominator is \(0\): \(x-2=0\), so \(x=2\). Near \(x=2\) the values blow up.
It has an infinite discontinuity (vertical asymptote) at \(x=2\).
Match the two pieces at \(x=2\): the left value must equal the right value.
| \(\lim_{x\to 2^{-}}f\) | \(=\) | \(3\cdot 2-1=5\) |
| \(f(2)\) | \(=\) | \(2+c\) |
| \(2+c\) | \(=\) | \(5\) |
| \(c\) | \(=\) | \(3\) |
Common pitfalls
Frequently asked questions
What does it mean for a function to be continuous?
Its graph has no breaks: at every point \(a\), the limit exists and equals \(f(a)\), so you can trace the curve without lifting your pen.
What are the three conditions for continuity at a point?
\(f(a)\) is defined, \(\lim_{x\to a} f(x)\) exists, and the two are equal. All three must hold.
What is a removable discontinuity?
A hole: the limit exists, but \(f(a)\) is missing or set to a different value. You could 'remove' it by plugging the hole with the limit value.
Where is a rational function discontinuous?
Exactly where its denominator equals \(0\). If that factor cancels with the numerator it is a removable hole; if not, it is an infinite discontinuity.
How do you make a piecewise function continuous?
Set the two pieces equal at the boundary point and solve for the unknown constant, so the left-hand value matches the right-hand value.