Learning Objectives
Check the three conditions for continuity at an interior point.
Choose a parameter that makes a piecewise function continuous.
Apply the Intermediate Value Theorem with an explicit continuity justification.
Prerequisites: one-sided limits and algebraic limit evaluation. This lesson connects local behavior near a point with a guarantee about values across an interval.
1. The Three Continuity Checks
A function f is continuous at an interior input a when f(a) is defined, lim(x → a) f(x) exists, and that limit equals f(a). These are separate checks. A graph with a hole fails the first check; a jump fails the second; a misplaced filled point can fail the third.
Check | Question |
|---|---|
Value | Is f(a) defined? |
Limit | Do the left and right sides approach the same finite height? |
Agreement | Is that height exactly f(a)? |
On a closed interval [a, b], continuity at the left endpoint uses the right-hand limit, and continuity at the right endpoint uses the left-hand limit. Polynomials are continuous everywhere. Rational functions are continuous wherever their denominators are nonzero.
2. Repair a Removable Discontinuity
Worked Example: Filling a Hole
Define f(x) = (x² − 1)/(x − 1) for x ≠ 1, and f(1) = k. Find k for continuity at 1.
For x ≠ 1, factor and simplify: f(x) = x + 1. Hence lim(x → 1) f(x) = 2.
Continuity requires f(1) to equal this limit, so k = 2. Changing a single point can repair a hole because the nearby behavior already agrees.
3. Join Two Pieces
Worked Example: Matching at a Boundary
Let g(x) = 2x + c for x < 3, and g(x) = x² − 1 for x ≥ 3. Find c for continuity at 3.
The left-hand limit is 6 + c. The right-hand limit and g(3) are both 9 − 1 = 8.
Set 6 + c = 8 to get c = 2. With this choice, both limits and the function value equal 8.
A genuine jump cannot be repaired by changing only the value at the boundary. When the one-sided limits differ, the nearby formulas themselves must change if continuity is required.
4. The Intermediate Value Theorem
If f is continuous on [a, b] and N lies strictly between f(a) and f(b), then there is at least one c in (a, b) with f(c) = N. The theorem guarantees existence. It does not locate every solution or guarantee that only one solution exists.
Worked Example: Guaranteeing a Root
Show that p(x) = x³ + x − 1 has a zero between 0 and 1.
A polynomial is continuous on [0, 1]. Its endpoint values are p(0) = −1 and p(1) = 1.
Since 0 lies between −1 and 1, the Intermediate Value Theorem guarantees at least one c in (0, 1) with p(c) = 0. No numerical root estimate is needed for this existence argument.
Common Mistakes
Do not cite the Intermediate Value Theorem without checking continuity on the whole interval.
Opposite signs alone are insufficient: 1/x is negative at −1 and positive at 1, but is undefined at 0 and never equals zero.
Do not conclude uniqueness from the Intermediate Value Theorem.
5. Practice
1. Let h(x) = (x² − 4)/(x − 2) for x ≠ 2 and h(2) = 9. Is h continuous at 2? Give all three checks.
2. Define q(x) = ax + 1 for x < 2 and q(x) = 7 for x ≥ 2. Find a for continuity.
3. Prove that x³ − 2 has a root in (1, 2).
Worked Solutions
1. h(2) is defined and equals 9. The limit exists and equals 4 after cancellation. Since 9 ≠ 4, h is not continuous. Redefining h(2) as 4 would repair it.
2. The left-hand limit is 2a + 1; the right-hand limit and q(2) are 7. Solve 2a + 1 = 7 to obtain a = 3.
3. The polynomial is continuous on [1, 2], with values −1 and 6 at the endpoints. Zero lies strictly between them, so the theorem guarantees a root in (1, 2).
6. Summary
- Continuity requires a value, a finite limit, and equality between them.
- At a piecewise boundary, compare both formulas and the included endpoint value.
- The Intermediate Value Theorem guarantees an intermediate output only when continuity holds throughout the interval.