Continuity
A function is continuous at when where it's heading matches where it actually is: no holes, jumps or blow-ups. You could draw it without lifting your pencil, so on an interval it can't skip over any value in between.
What the symbols mean
- The value at a
Must exist and match the limit.
What the symbols mean
- Target value
Any number between the endpoint values gets hit somewhere in between.
Explanation
A continuous function is simply one whose graph has no breaks, so you can draw it without lifting your pencil. It matters because continuous functions behave predictably: small changes in the input give small changes in the output, and plugging in gives the limit. The definition is just two things being equal: (where the function is heading as you approach ) and (where it actually is when you get there). The Intermediate Value Theorem looks long, but it says something you already believe: if a pencil line starts below some height and ends above it, it crossed that height somewhere. Bear with the symbols; the ideas are simple.
Every lesson so far has made a point of separating where is heading from what actually is. Continuity is the happy case where those two match. That takes three things: is defined, exists, and they're equal.
Kinds of discontinuity. Each one is a way those three can fail, and you've met all of them already:
- Removable: a hole, like at . The limit exists but is missing or wrong. Redefining that one point fixes it.
- Jump: the one-sided limits differ.
- Infinite: a vertical asymptote.
Continuous everywhere on their domains: polynomials, rational functions, roots, trig functions, , , and their sums, products, quotients and compositions. This is really the limit laws in disguise: they told us the limit of a polynomial is its value, and that's exactly what continuous means. It's also why plugging in has worked so often.
The Intermediate Value Theorem is where continuity starts paying off: it guarantees solutions exist. Suppose is continuous, and . The graph starts below the -axis and ends above it, and since it has no jumps, it can't get from one side to the other without crossing. So for some between 1 and 2. The theorem doesn't say where, but you can hunt the root down by checking the midpoint and keeping whichever half still has a sign change. That's the bisection method.
See it
Worked examples
Show that has a solution between 1 and 2.
- 1
is a polynomial, so it's continuous and the IVT applies.
Practice
Type your answer and check it. Solve them all to complete the lesson.
Find so that is continuous.
Where is discontinuous?
has a hole at . What value of makes it continuous?
Does have a solution in ?
Find the root of in to three decimals (bisect, or use a calculator).
A badly written tax law charges of your whole income if you earn up to \40{,}00020%$40{,}000$?
The hardest one here. Get it, and you've got this lesson.
Find and so that is continuous everywhere.
Where it's used
The IVT is why bisection root-finders never fail on a sign change, and continuity is the formal version of "small changes in input cause small changes in output," which every stable control system depends on.