Continuity
In short: Continuity at a point means the graph has no break there: the function is defined, its limit exists, and the two are equal, so the curve can be drawn without lifting the pencil.
AP Calculus ABCalculusAP Calculus AB Unit 1
What is Continuity?
The idea of continuity is a graph you can trace in one stroke. At an ordinary point, the outputs nearby line up exactly with the output at the point itself, with no hole, gap or jump in the way.
A function \(f\) is continuous at \(x = c\) when three things hold: (1) \(f(c)\) is defined; (2) \(\lim_{x \to c} f(x)\) exists; and (3) \(\lim_{x \to c} f(x) = f(c)\). If any step fails, the function is discontinuous at \(c\), whether through a removable hole, a jump or an infinite break.
How to spot it
- A graph is drawn with a hollow circle, a jump between pieces, or a vertical asymptote, and the question asks about continuity there.
- A piecewise function has a joint at which both formulas must meet at the same height.
- A problem asks for the value of a constant that makes a function connect smoothly at a point.
- A theorem such as the Intermediate Value Theorem requires the function to be continuous on an interval.
The method
- To test continuity, check condition 1: substitute \(c\) and see whether \(f(c)\) exists.
- Check condition 2: compute the left and right approach values and see whether they agree.
- Check condition 3: compare the approach value with \(f(c)\).
- Name the break if one fails: removable (hole), jump (two sides differ) or infinite (vertical asymptote).
Continuity: solved example
Problem
Let \(f(x) = \dfrac{x^2 – 4}{x – 2}\) for \(x \neq 2\) and \(f(2) = 5\). Is \(f\) continuous at \(x = 2\)? If not, how could \(f(2)\) be redefined to repair it?
Solution
Condition 1 holds: \(f(2) = 5\). For condition 2, \(\frac{x^2 – 4}{x – 2} = x + 2\) when \(x \neq 2\), so the approach value is \(2 + 2 = 4\).
Since \(4 \neq 5\), condition 3 fails: \(f\) is not continuous at 2. Setting \(f(2) = 4\) removes the break.
Try it
Question 1
Is \(g(x) = \frac{1}{x – 3}\) continuous at \(x = 3\)?
Show the answer
No. \(g(3)\) is undefined, so condition 1 fails; the graph has a vertical asymptote there.
Question 2
Find \(k\) so that \(h(x) = 3x + k\) for \(x < 2\) and \(h(x) = x^2 - 1\) for \(x \ge 2\) is continuous at 2.
Show the answer
Match the pieces: \(6 + k = 3\), so \(k = -3\).
Continuity vs. Limit
A limit looks only at the neighborhood and says where the outputs are headed. This property adds a requirement about the point itself: the function must exist there and sit at that same height. So a continuous function always has a matching limit, but a limit can exist where the function is broken.
Keep going
Lessons
Worksheets and quizzes
- Continuity and the Intermediate Value Theorem – Worksheet
- Continuity and the Intermediate Value Theorem – Quiz
Related words
Whole dictionary: A–Z · AP Calculus AB · Calculus