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: Picture It
Start with something you already know.
Crossing a footbridge
- A footbridge is built in two halves that meet over a stream.
- To walk straight across, there has to be a plank at the meeting spot.
- Both halves also have to arrive at the same height.
- And the plank has to sit at exactly that height, not above or below it.
- Miss any one of the three and you cannot walk across without a step or a jump.
Continuity at a point asks the same three things.
Is there a value, do both sides head for one height, and is the value at that height?
A missing plank is a hole. A plank at the wrong height can be moved, so it is removable too. Halves at different heights are a jump, and a half that climbs forever is an asymptote.
Where the picture breaks: on a real bridge you can step over a tiny gap, but a function is broken even if one single point is missing.
Now the same three checks with numbers, for x + 3 at x = 2.
| The check | For x + 3 at x = 2 |
|---|---|
| Is there a value at 2? | Yes: 2 + 3 = 5 |
| Do both sides head for one number? | Yes: 1.99 gives 4.99, 2.01 gives 5.01 |
| Is the value that number? | Yes: 5 and 5 |
All three checks pass, so x + 3 is continuous at 2. Every plank is in place.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
f(x) = (x² − 1)/(x − 1) for every x except 1, and f(1) = 5. Find where f is heading as x approaches 1, then compare it with f(1). Which sentence fits?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
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 3: 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 4: 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 5: Watch an Example
One step at a time.
Watch Maya Test a Point and Repair It
h(x) = (x² + x − 6)/(x − 2). Maya asks whether h is continuous at x = 2, and if not, what kind of break it has.
- Step 1
Condition one asks for a value at 2. Substituting gives (4 + 2 − 6)/(2 − 2) = 0/0, which is not a number, so h(2) does not exist. Condition one fails, and h is not continuous at 2.
Watch Leo Join Two Pieces, Then Finish It Yourself
f(x) = 3x + 1 for x < 2, and f(x) = x² + k for x ≥ 2. Leo wants the number k that makes f continuous at 2. He works the left side; the right side, k, and what goes wrong with another k are yours.
- Step 1
The rule changes at x = 2, so the two sides have to be asked separately. Below 2 the function is 3x + 1; at 2 and above it is x² + k.
Watch Ana Trap a Root
Ana wants to know whether f(x) = x³ + x − 3 has a zero between 1 and 2, without solving f(x) = 0.
- Step 1
Ana works out the ends. At x = 1 the value is 1 + 1 − 3 = −1. At x = 2 it is 8 + 2 − 3 = 7. Zero sits between −1 and 7.
Step 6: Your Turn
Practice makes it stick.
The Bulk Coffee Price
Problem 1 of 14
A store sells coffee by weight. Under 5 pounds it costs $12 a pound, so w pounds cost 12w dollars. At 5 pounds or more it costs $10 a pound plus a fixed charge of k dollars, so 10w + k. What fixed charge k makes the price continuous at 5 pounds, with no sudden jump?
dollars
The Cooling Bath
Problem 2 of 14
A bath is 104 °F at 7:00 and 86 °F at 7:30. Its temperature changes continuously. Which statement can be justified from this alone?
Continuous or Not?
Test the three conditions in order: the value, the limit, and whether they agree.
Problem 3 of 14
p(x) = (x² − 36)/(x − 6) for every x except 6, and p(6) = 12. How many of the three continuity conditions hold at x = 6?
Problem 4 of 14
q(x) = (x² − 2x − 8)/(x − 4) has a hole at x = 4. What value of q(4) would make q continuous there?
Problem 5 of 14
g(x) = 2x for x < 3, and g(x) = x + 5 for x ≥ 3. What happens at x = 3?
Problem 6 of 14
At how many values of x is r(x) = (x + 2)/((x − 1)(x + 5)) discontinuous?
Problem 7 of 14
h(x) = kx for x < 2, and h(x) = x² + 4 for x ≥ 2. What number k makes h continuous at 2?
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 14
Ravi wanted the k that makes f(x) = x² + 3 for x < 2 and f(x) = 3x + k for x ≥ 2 continuous at 2. One line is a slip. Which?
Ravi’s working for k at x = 2
- 1left limit: 2² + 3 = 7
- 2right limit and value: 3(2) + k = 6 + k
- 37 = 6 + k
- Answerk = 7 + 6 = 13
Problem 9 of 14
Solved correctly, what number k makes Ravi’s f continuous at 2?
Problem 10 of 14
Noor tested g(x) = (x² + 2x − 3)/(x − 1) for x ≠ 1, with g(1) = 2, at x = 1. One line is a slip. Which?
Noor’s test of g at x = 1
- 1g(1) = 2: condition one holds
- 2(x − 1)(x + 3)/(x − 1) = x + 3, for x ≠ 1
- 3limit = 1 + 3 = 4: condition two holds
- Answerboth exist, so g is continuous at 1
Problem 11 of 14
With all three conditions tested, which describes Noor’s g at x = 1?
Pick the Move
Choose the move that settles the question. Neither needs working all the way through.
Problem 12 of 14
Sam says 1/(x − 1) is −1 at x = 0 and 1 at x = 2, so by the Intermediate Value Theorem it equals 0 somewhere between. Does his argument hold?
Problem 13 of 14
You want every x where r(x) = (x − 2)/(x² − 4) is discontinuous. What is the first move?
Same Kind of Break?
Judge each one without drawing it. Select every entry that fits.
Problem 14 of 14
Select every function whose break at x = 2 is removable.
Step 7: Quick Check
Show what you know.
Question 1 of 1
f(4) = 1 and the limit at 4 is 6. Continuous?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
f(x) = (x² + bx + 6)/(x − 2) for x ≠ 2, and f(2) = c. For exactly one b the fraction has a limit at 2, and then one c makes f continuous there. Find that c.
Stretch 2 of 2
f is continuous on [0, 4], with f(0) = 1, f(2) = −3 and f(4) = 1. What is the smallest number of zeros f can have strictly between 0 and 4?
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.
For the grown-up
The misconception this lesson targets is that continuity means “the function has a value there”, or that a value and a limit existing side by side is enough. A student holding it calls a filled-in hole at the wrong height continuous, because both numbers are present, and never compares them. It travels with a second habit: reading the Intermediate Value Theorem as a fact about end values alone, and applying it across an asymptote. The fix is the same both times: say the three conditions out loud in order, value, limit, agreement, and before any theorem, state that its function is continuous on the closed interval.