1.11 Defining Continuity at a Point: Calculus Guide (2025)
A function is continuous at a point when three conditions hold at that same input value: the function is defined there, the limit exists there, and the limit equals the function’s value. Formally, f is continuous at x = a if lim(x→a) f(x) = f(a). This single equation in 1.11 defining continuity at a point bundles three separate requirements, which is why most exam errors come from checking only one or two of them.
Definition & Conditions
The formal definition states that a function f is continuous at x = a if and only if all three of the following are true:
- f(a) is defined. The point a must be in the domain of f, so there is an actual output value to compare against.
- lim(x→a) f(x) exists. The two-sided limit must approach a single finite number. That means the left-hand limit and right-hand limit both exist and are equal.
- The limit equals the value. lim(x→a) f(x) = f(a). The limit and the function value must match, not merely both exist.
Each condition fails in a different way. A hole in the graph breaks condition one. A vertical asymptote or a jump breaks condition two. A removable hole where the limit exists but the function is defined differently breaks condition three. Continuity requires all three simultaneously, which is a stricter demand than the existence of a limit alone.
A function is continuous on an open interval if it is continuous at every point inside it. Continuity at a single point is the building block: intervals, and later theorems like the Intermediate Value Theorem, are assembled from point-by-point continuity. Khan Academy’s limits and continuity unit treats the three-part definition as the foundation for everything that follows, which is why the pointwise version deserves careful attention before moving on.
Types of Discontinuities
When at least one condition fails, the point is a discontinuity. Discontinuities fall into three standard categories.
- Removable discontinuity. The limit exists, but it does not equal f(a), or f(a) is undefined. Example: f(x) = (x² − 1)/(x − 1) at x = 1. The limit is 2, but f(1) is undefined because of division by zero. The graph is a line with a single hole.
- Jump discontinuity. The left-hand and right-hand limits both exist but differ, so the two-sided limit does not exist. Example: a step function that outputs 0 for x < 0 and 1 for x ≥ 0 has a jump at x = 0.
- Infinite discontinuity. At least one one-sided limit is infinite, producing a vertical asymptote. Example: f(x) = 1/x at x = 0. Neither one-sided limit is finite, so the function cannot be continuous there.
The label “removable” is literal. If the discontinuity is removable, you can redefine a single point — set f(a) equal to the limit — and the function becomes continuous. Jump and infinite discontinuities cannot be repaired this way because no finite value could match a limit that does not exist. Recognizing which type you face tells you immediately whether the problem asks for a redefinition or simply a classification.
Testing Continuity Steps
A reliable routine prevents skipped conditions. Apply these steps in order at the given point a.
- Check the domain. Confirm f(a) is defined. If a is excluded from the domain, the function is discontinuous there and you can stop.
- Compute the left-hand limit. Evaluate lim(x→a⁻) f(x). For piecewise functions, use the rule that applies for x < a.
- Compute the right-hand limit. Evaluate lim(x→a⁺) f(x) using the rule for x > a.
- Compare the one-sided limits. If they differ, the two-sided limit does not exist — the discontinuity is a jump.
- Compare the limit to the value. If lim(x→a) f(x) = f(a), the function is continuous at a. If the limit exists but differs from f(a), the discontinuity is removable.
Piecewise definitions are the most common exam setting for this test because each condition can be checked against a different rule. The left-hand limit uses one branch, the right-hand limit uses another, and the value uses whichever branch contains a itself. MIT OpenCourseWare’s single-variable calculus materials build piecewise continuity problems around exactly this split, and it is worth practicing the branch-by-branch approach until it is automatic.
Worked Example
Solving for a constant in a piecewise function. Let f(x) = 3x + k for x < 2, and f(x) = x² + 1 for x ≥ 2. Find k so that f is continuous at x = 2.
- Left-hand limit: lim(x→2⁻) (3x + k) = 6 + k.
- Right-hand limit and value: lim(x→2⁺) (x² + 1) = 4 + 1 = 5, and f(2) = 5.
- Set them equal: 6 + k = 5, so k = −1.
With k = −1, the left-hand limit is 5, matching both the right-hand limit and f(2), so the function is continuous at x = 2.
Quick Reference Table
| Discontinuity | Left & right limits | Does limit = f(a)? | Repairable by redefining one point? |
|---|---|---|---|
| Removable | Equal and finite | No (f(a) missing or different) | Yes |
| Jump | Unequal but both finite | No (limit doesn’t exist) | No |
| Infinite | At least one is ±∞ | No | No |
Key Takeaways
- A function is continuous at x = a only if f(a) is defined, the two-sided limit exists, and the two are equal — all three, not any one.
- The condition can be written compactly as lim(x→a) f(x) = f(a), which is the standard test.
- Removable discontinuities have matching one-sided limits but a mismatched or missing value, and can be fixed by redefining a single point.
- Jump discontinuities have unequal finite one-sided limits; infinite discontinuities involve a vertical asymptote. Neither can be repaired.
- For piecewise functions, compute the left limit, the right limit, and the value separately, then equate them to solve for unknown constants.
- Polynomials and most standard functions are continuous on their entire domains, so focus the full test on boundaries, division points, and roots.
