No breaks, no jumps
Informally, a function is continuous if you can draw its graph without lifting your pencil.
The formal test
f is continuous at x = a if all three are true:
- f(a) is defined.
- lim (x→a) f(x) exists.
- lim (x→a) f(x) = f(a).
If any condition fails, f is discontinuous at a.
Types of discontinuity
| Type | What happens | Example |
|---|---|---|
| Removable discontinuity (hole) | the limit exists but f(a) is missing or different | (x² − 1)/(x − 1) at x = 1 |
| Jump discontinuity | left and right limits differ | a step function, like parking fees |
| Infinite discontinuity | the function goes to ±∞ | 1/x at x = 0 |
Checking a piecewise function
f(x) = { x² for x < 2; 3x − 2 for x ≥ 2 }
- f(2) = 3(2) − 2 = 4 ✓
- Left limit: 2² = 4; right limit: 4 → limit = 4 ✓
- lim = f(2) ✓
Continuous at x = 2.
Finding a value that makes a function continuous
f(x) = { kx + 1 for x < 3; x² − 2 for x ≥ 3 }
- Right side at 3: 9 − 2 = 7
- Left side at 3: 3k + 1
- Set equal: 3k + 1 = 7 → k = 2
Continuous functions to know
| Function | Continuous where? |
|---|---|
| polynomials | everywhere |
| exponential functions | everywhere |
| sin x, cos x | everywhere |
| rational functions | everywhere except where the denominator is 0 |
| √x | x ≥ 0 |
| log x | x > 0 |
Sums, products and compositions of continuous functions are continuous (where defined).
The Intermediate Value Theorem
If f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b).
Use: proving roots exist. f(x) = x³ − x − 1 has f(1) = −1 and f(2) = 5. Since f is continuous and changes sign, there’s a root between 1 and 2.
Real-world continuity
- Temperature changes continuously over time.
- Cost of postage jumps at weight cut-offs (jump discontinuity).
- Population counts are technically whole numbers but are often modelled as continuous.
Worked examples
Example 1: Is f(x) = 1/(x − 4) continuous at x = 4? No: infinite discontinuity.
Example 2: Type of discontinuity of (x² − 9)/(x − 3) at x = 3? Removable (hole at (3, 6)).
Example 3: Does x³ + x − 3 have a root between 0 and 2? f(0) = −3, f(2) = 7 → yes (IVT).
Common mistakes
- Checking only whether f(a) exists.
- Ignoring one-sided limits for piecewise functions.
- Applying the IVT to discontinuous functions.
Try it yourself
- Where is f(x) = (x + 1)/(x² − 4) discontinuous?
- Classify the discontinuity of f(x) = |x|/x at x = 0.
- Find k so that f(x) = { 2x + k, x < 1; 5, x ≥ 1 } is continuous.
- Is √x continuous at x = 4?
Answers: 1) x = 2 and x = −2 2) jump (left −1, right 1) 3) k = 3 4) yes
Quick recap
- Continuity at a: f(a) defined, the limit exists, and they’re equal.
- Types: removable discontinuity, jump discontinuity, infinite discontinuity.
- The Intermediate Value Theorem guarantees in-between values for continuous functions.