Calculus I Foundations (MATH 151) · Module 2 of 11
Continuity & the Intermediate Value Theorem
The precise definition of continuity, the ways a function can fail it, and a powerful existence theorem.
Defining Continuity
- f is continuous at x = a if three conditions hold: f(a) is defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a).
A function has a hole at x = 2 (f(2) is undefined) but the limit as x approaches 2 exists. Is the function continuous at x = 2?
Solution. No — continuity requires f(2) to be defined; since it isn't, the function is discontinuous at x = 2, specifically a removable discontinuity.
Types of Discontinuity
- A removable discontinuity (a hole) occurs when the limit exists but doesn't match f(a), or f(a) is undefined; it can be 'fixed' by redefining f(a).
- A jump discontinuity occurs when the left and right limits exist but differ; an infinite discontinuity occurs at a vertical asymptote.
What distinguishes a jump discontinuity from a removable discontinuity?
Solution. A jump discontinuity occurs when the left-hand and right-hand limits both exist but differ from each other; a removable discontinuity occurs when the two-sided limit does exist, but it doesn't match the function's actual value there, or the function is undefined there — and can be fixed by redefining that one point.
The Intermediate Value Theorem (IVT)
- The IVT states that if f is continuous on [a,b] and N is any value between f(a) and f(b), then there exists at least one c in [a,b] where f(c) = N.
A continuous function has f(1) = −3 and f(4) = 5. Does the function have a root, a value where f(x) = 0, somewhere between x = 1 and x = 4?
Solution. Yes — by the Intermediate Value Theorem, since 0 lies between f(1) = −3 and f(4) = 5, and f is continuous on [1,4], there must be at least one c in [1,4] where f(c) = 0.
Continuity of Common Functions
- Polynomials are continuous everywhere; rational functions are continuous everywhere except where the denominator equals zero; trigonometric functions like sine and cosine are continuous everywhere.
Where is the function f(x) = 1/(x − 5) discontinuous?
Solution. At x = 5, where the denominator equals zero, making the function undefined there — an infinite discontinuity and vertical asymptote.
Check yourself
2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.
1. What three conditions define continuity at x = a?
2. What type of discontinuity occurs when the left and right limits exist but differ?