12.3Continuity
Verify continuity using the three-condition test, classify removable, jump, and infinite discontinuities, and apply the Intermediate Value Theorem to guarantee the existence of roots and solutions.
Continuity is the bridge between limits and derivatives. A function must be continuous at a point before it can be differentiable there. The Intermediate Value Theorem is one of the most powerful tools in analysis, guaranteeing solutions exist even when we cannot find them explicitly.
What Is Continuity?
A function is continuous at a point x = c when its graph has no holes, jumps, or vertical asymptotes at that point — you can draw through it without lifting your pencil. Formally, three conditions must all hold simultaneously.
Three Conditions for Continuity at x = c
- f(c) is defined — the function value exists.
- limx→c f(x) exists — the two-sided limit exists.
- limx→c f(x) = f(c) — the limit equals the function value.
If any one condition fails, the function is discontinuous at x = c.
Types of Discontinuity
Removable Discontinuity
The limit exists but either f(c) is undefined or f(c) ≠ lim f(x). The graph has a 'hole' that could be 'filled in' to make the function continuous.
Example: f(x) = (x²−4)/(x−2) has a hole at x = 2 because the factor (x−2) cancels.
Jump Discontinuity
The left-hand and right-hand limits both exist but are not equal. The graph 'jumps' from one value to another.
Example: A piecewise function where the two pieces approach different values at the boundary.
Infinite Discontinuity
The function approaches ±∞ as x → c, producing a vertical asymptote. The limit does not exist (as a finite number).
Example: f(x) = 1/x has an infinite discontinuity at x = 0.
Continuity on an Interval
A function is continuous on an open interval (a, b) if it is continuous at every point in that interval. It is continuous on a closed interval [a, b] if it is continuous on (a, b) and the one-sided limits match the function values at the endpoints.
Intermediate Value Theorem (IVT)
Theorem
If f is continuous on the closed interval [a, b] and N is any number strictly between f(a) and f(b), then there exists at least one number c in (a, b) such that f(c) = N.
Consequence: if f(a) and f(b) have opposite signs, then f has at least one root (zero) in (a, b).
Worked Examples
Verify that f(x) = x² − 1 is continuous at x = 2 using the three-condition test.
Condition 1 — f(2) exists: f(2) = (2)² − 1 = 4 − 1 = 3. ✓
Condition 2 — lim(x→2) (x² − 1) exists: the function is a polynomial, so the limit = 3. ✓
Condition 3 — lim(x→2) f(x) = f(2): 3 = 3. ✓
All three conditions hold.
Identify and classify the discontinuity of f(x) = (x² − 4)/(x − 2) at x = 2.
Condition 1 — f(2): denominator = 0, so f(2) is undefined. ✗
Condition 2 — lim(x→2): factor the numerator: (x²−4)/(x−2) = (x+2)(x−2)/(x−2) = x+2 for x ≠ 2.
So lim(x→2) f(x) = 2 + 2 = 4. The limit exists.
Because the limit exists but f(2) is undefined, this is a removable discontinuity.
Determine if f(x) = { x + 1 if x < 3 ; 2x − 2 if x ≥ 3 } is continuous at x = 3.
Condition 1 — f(3): use x ≥ 3 piece: f(3) = 2(3) − 2 = 4. Defined. ✓
Condition 2 — left-hand limit: lim(x→3⁻) (x+1) = 3+1 = 4.
Right-hand limit: lim(x→3⁺) (2x−2) = 2(3)−2 = 4.
Both one-sided limits equal 4, so lim(x→3) f(x) = 4. ✓
Condition 3 — lim = f(3): 4 = 4. ✓
Find the value of k that makes f(x) = { kx + 1 if x ≤ 2 ; x² − 1 if x > 2 } continuous at x = 2.
For continuity, lim(x→2⁻) f(x) must equal lim(x→2⁺) f(x) = f(2).
Left-hand limit: lim(x→2⁻) (kx+1) = 2k + 1.
Right-hand limit: lim(x→2⁺) (x²−1) = 4 − 1 = 3.
f(2) = k(2) + 1 = 2k + 1 (using the x ≤ 2 piece).
Set left limit = right limit: 2k + 1 = 3 → 2k = 2 → k = 1.
Use the IVT to show that f(x) = x³ − x − 1 has a root on [1, 2].
f is a polynomial, so it is continuous on [1, 2]. ✓
f(1) = 1 − 1 − 1 = −1. (negative)
f(2) = 8 − 2 − 1 = 5. (positive)
Since f(1) = −1 {'<'} 0 {'<'} 5 = f(2), by the IVT there exists c ∈ (1, 2) with f(c) = 0.
Guided Practice
Check the continuity of f(x) = (x + 3)/(x − 1) at x = 1 and at x = 3.
Hint: At x = 1, check whether the denominator is zero. At x = 3, evaluate all three conditions.
Classify the discontinuity of f(x) = |x|/x at x = 0.
Hint: Compute the left-hand limit (x → 0⁻) and the right-hand limit (x → 0⁺) separately. Do they agree?
Determine whether f(x) = { 2x − 1 if x ≤ 1 ; x² if x {'>'} 1 } is continuous at x = 1.
Hint: Compute f(1), lim(x→1⁻), and lim(x→1⁺). Check all three conditions.
Find k so that f(x) = { 3x − k if x {'<'} 2 ; kx² + 1 if x ≥ 2 } is continuous at x = 2.
Hint: Set the left-hand limit equal to f(2) and solve for k.
Use the IVT to confirm that g(x) = x⁴ − 3x + 1 has a root on [0, 1].
Hint: Evaluate g(0) and g(1). Are they opposite in sign? Is g continuous on [0, 1]?
Common Mistakes
Assuming a function is continuous just because the limit exists.
The limit must exist AND equal f(c), and f(c) must be defined. All three conditions are required.
Confusing removable and jump discontinuities — both have a 'break' in the graph.
Removable: the limit exists (both sides agree) but ≠ f(c) or f(c) undefined. Jump: left and right limits exist but differ.
Applying the IVT without verifying continuity on the interval first.
The IVT requires f to be continuous on [a, b]. Always state this as the first step.
Thinking the IVT tells you exactly where the root is.
The IVT only guarantees existence of a root — it does not find the root's location.
Math Tips
Polynomials and rational functions are continuous everywhere they are defined. Rational functions are discontinuous only where the denominator equals zero.
To find a removable discontinuity, factor and cancel. The cancelled factor reveals the x-value of the hole.
For piecewise functions, always check continuity at the boundary points — that is where discontinuities can occur.
The IVT is an existence theorem. Think of it as: 'if you start below zero and end above zero on a continuous path, you must cross zero somewhere.'
Interactive Practice — 5 Questions
Which of the following is NOT one of the three conditions required for f to be continuous at x = c?
f(x) = (x² − 9)/(x − 3) has what type of discontinuity at x = 3?
For f(x) = { x² if x ≤ 1 ; 2x − 1 if x > 1 }, what is true at x = 1?
The Intermediate Value Theorem guarantees a root of f on (a, b) when:
A function has a jump discontinuity at x = c. Which statement is true?