Continuity and the Intermediate Value Theorem
Formally define continuity using three conditions and classify all types of discontinuity. State and apply the IVT to guarantee the existence of roots and specific output values on closed intervals.
A lesson of AP Calculus AB, in Limits and Continuity, about 10 minutes.
By the end of this lesson, you will be able to define continuity precisely, identify types of discontinuities, and apply the Intermediate Value Theorem to prove the existence of solutions.
The Definition of Continuity
A function f is continuous at x = a if all three conditions hold:
- f(a) exists (the function is defined at a)
- lim[x→a] f(x) exists (the limit exists)
- lim[x→a] f(x) = f(a) (the limit equals the function value)
Intuitive meaning: There's no hole, jump, or break at x = a; you can draw the function through that point without lifting your pencil.
If ANY of the three conditions fails, f is discontinuous at a.
Types of Discontinuities
Removable Discontinuity (Hole)
The limit exists but either f(a) is undefined or f(a) ≠ the limit.
Example: f(x) = (x² - 4)/(x - 2) = (x-2)(x+2)/(x-2) = x + 2 for x ≠ 2.
At x = 2: f(2) is undefined (denominator = 0).
But: lim[x→2] f(x) = 4.
This is a removable discontinuity at x = 2 — there's a "hole" in the graph at (2, 4). It can be "removed" by defining f(2) = 4.
Jump Discontinuity
The left and right limits both exist but are different.
Example: f(x) = x/|x| (the sign function):
- For x > 0: f(x) = 1
- For x < 0: f(x) = -1
lim[x→0⁺] f(x) = 1, lim[x→0⁻] f(x) = -1. These are different, so the two-sided limit doesn't exist. Jump discontinuity at x = 0.
Infinite Discontinuity
The function approaches ±∞ at the discontinuity — a vertical asymptote.
Example: f(x) = 1/x at x = 0.
lim[x→0⁺] 1/x = +∞, lim[x→0⁻] 1/x = -∞.
Infinite discontinuity at x = 0.
Continuity on an Interval
f is continuous on an open interval (a,b) if it's continuous at every point in the interval.
f is continuous on a closed interval [a,b] if:
- It's continuous on (a,b)