A function is continuous at a if f(a) exists, the limit exists, and the two are equal.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Three ways to fail
A hole, a jump, or a vertical asymptote. Each breaks a different one of the three conditions.
Removable
A hole is removable: redefining a single point repairs it. A jump is not.
The informal version
You can draw it without lifting the pen — true, but it is the three conditions that get tested.
The three conditions
f(a) exists, the limit as x approaches a exists, and the two are equal. All three are required, and an exam justification should name whichever one fails rather than simply asserting discontinuity.
Piecewise functions are where it is tested
Finding a constant that makes a piecewise function continuous means setting the one-sided limits equal at the boundary. That is the standard question, and it reduces to solving one equation.
Continuity licenses the theorems
The intermediate value theorem, extreme value theorem and mean value theorem all require continuity. Citing a theorem without stating that its hypothesis holds is an incomplete justification on the exam.
Differentiable implies continuous
A function with a derivative at a point must be continuous there. The converse is false — |x| is continuous at zero and has no derivative. The implication runs one way only.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Nadia Test a Point
Nadia checks continuity of f at x = 2, where f(2) = 5 and the limit is 3.
- Step 1
Condition one: f(2) exists. It is 5, so that passes.
Step 4: Your Turn
Practice makes it stick.
The Three Conditions
Problem 1 of 2
How many conditions must hold for continuity at a point?
The Jump
Problem 2 of 2
Left limit 2, right limit 7. Is this discontinuity removable? 1 for yes, 0 for no.
Continuous or Not
1 of 8
f(3) = 4 and the limit at 3 is 4. Continuous? 1 for yes, 0 for no.
2 of 8
f(3) undefined, limit is 4. Continuous? 1 for yes, 0 for no.
3 of 8
Sort each discontinuity by whether redefining one point repairs it.
Tap something to move it.
- Empty
- Empty
4 of 8
Is a polynomial continuous everywhere? 1 for yes, 0 for no.
5 of 8
1/(x − 6). At which x is it discontinuous?
6 of 8
A hole at x = 2 with limit 9. What value at 2 would repair it?
7 of 8
Left limit 5, right limit 5, f(1) = 5. Continuous at 1? 1 for yes, 0 for no.
8 of 8
How many conditions fail at a vertical asymptote where f is undefined?
Step 5: Quick Check
Show what you know.
Question 1 of 1
f(4) = 1 and the limit at 4 is 6. Continuous? 1 for yes, 0 for no.
What You Learned
- Continuity at a point needs the value, the limit, and their agreement.
- A hole is removable; a jump and an asymptote are not.
- Polynomials are continuous everywhere.