A function is continuous at a if f(a) exists, the limit exists, and the two are equal. Three ways to fail: a hole, a jump, or a vertical asymptote. Each breaks a different condition, and only the hole is removable.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Each condition alone
(x² − 4)/(x − 2) has a limit at 2 but no value: a hole. A step function has a value but no limit: a jump. 1/(x − 2) has neither: an asymptote.
Removable
Redefining a single point repairs a hole: set f(2) = 4 and (x² − 4)/(x − 2) becomes x + 2 everywhere. A jump cannot be repaired by one point; the two sides disagree.
Making a join continuous
f(x) = x² for x < 1 and f(x) = 2x + k for x ≥ 1: the pieces meet only if 1 = 2 + k, so k = −1. Set the one-sided limits equal and solve.
Continuous families
Polynomials are continuous everywhere. So are sine, cosine and exponentials. Rational functions are continuous wherever the denominator is not zero. Roots are continuous on their domains.
The informal version
You can draw it without lifting the pen. True, but it is the three conditions that get tested, and the informal version cannot tell a hole from a jump.
What continuity buys
The Intermediate Value Theorem: a continuous f on [a, b] takes every value between f(a) and f(b). x³ − x − 1 goes from −1 at x = 1 to 5 at x = 2, so it has a root between. Continuity is the hypothesis that does the work.
Step 2: Try It Yourself
Tap and try it out.
f(x) = x² for x < 1 and f(x) = 2x + k for x ≥ 1. Find k for continuity at 1, then classify the break when k = 0
- 1left limit: 1² = 1x → 1⁻
Step 0 of 4
Step 3: In Real Life
A thermostat
Room temperature changes continuously; it cannot jump from 68 to 72 without passing 70. A control system assumes that, and a sensor reading that jumps signals a fault, not a real change.
Step 4: 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 5: 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?
Continuous or Not
1 of 8
f(3) = 4 and the limit at 3 is 4. Continuous?
2 of 8
f(3) undefined, limit is 4. Continuous?
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?
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?
8 of 8
How many conditions fail at a vertical asymptote where f is undefined?
Step 6: Quick Check
Show what you know.
Question 1 of 1
f(4) = 1 and the limit at 4 is 6. Continuous?
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.
- To join two pieces, set the one-sided limits equal and solve.
- Polynomials are continuous everywhere, and continuity is what the Intermediate Value Theorem needs.