Historical Context & Motivation
The concept of continuity—and the problems that arise when it fails—has driven some of the most important developments in the history of calculus. Long before rigorous definitions existed, mathematicians recognized that certain expressions break down at particular points even though the overall behavior of the function seems perfectly well-behaved nearby. The desire to "repair" these breakdowns motivated centuries of work on limits, function definition, and the very foundations of analysis. Understanding how and when a removable discontinuity can be eliminated by a simple redefinition reveals a deep principle: the value a function approaches matters more than the value it happens to take.
The central question this lesson addresses is deceptively simple: if a function is undefined or incorrectly defined at a single point, yet the limit exists there, can we patch the hole? The answer is yes—and the technique for doing so is both elegant and heavily tested on the AP Calculus AB exam. Mastering it requires fluency with limits, factoring, and the formal definition of continuity, all of which converge in this single topic.
Core Principles & Definitions
Before we can remove a discontinuity, we need a precise vocabulary for what continuity means and how it can fail. Recall that a function f is continuous at x = a if and only if three conditions hold simultaneously: f(a) is defined, lim as x→a of f(x) exists, and lim as x→a of f(x) equals f(a). A removable discontinuity arises when the second condition is met—the limit exists—but either the first or third condition fails. This is the only type of discontinuity that can be "removed" by redefining the function at a single point.
Continuity at a Point
Removable Discontinuity
Non-Removable Discontinuity
Removing the Discontinuity
Visual Explanation
A graph is the most immediate way to see the difference between a removable and a non-removable discontinuity. The diagram below shows a function with a removable discontinuity (a hole) at x = 2 alongside the repaired version of the function. Notice that the curve is identical on both sides of x = 2; only the single point is missing in the original version. The open circle indicates the hole, and the filled circle shows the value that restores continuity.
The key geometric insight is that the curve passes through the point (2, 4) in a limiting sense—both the left-hand and right-hand limits converge there—but the function either skips or misassigns that y-value. The algebraic simplification (x² − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2 for x ≠ 2 reveals the continuous function hiding behind the fraction. The cancellation of the common factor is the algebraic act of "removing" the discontinuity.
Mathematical Framework
The formal procedure for removing a discontinuity rests on the definition of continuity and the evaluation of limits. The process has three stages: identify the point of discontinuity, compute the limit, and redefine the function. Each stage connects to specific algebraic techniques that appear repeatedly on the AP exam.
Classifying Discontinuities
Not every discontinuity can be removed, so a critical skill on the AP exam is classifying the type of discontinuity present. The three principal types—removable, jump, and infinite—are distinguished by the behavior of the one-sided limits. The diagram below places all three side by side for comparison, making the visual signatures unmistakable.
| Property | Removable | Jump | Infinite |
|---|---|---|---|
| lim(x→a⁻) f(x) | = L (finite) | = L₁ (finite) | = ±∞ |
| lim(x→a⁺) f(x) | = L (same) | = L₂ ≠ L₁ | = ±∞ |
| Two-sided limit | Exists (= L) | Does not exist | Does not exist |
| Can be removed? | Yes | No | No |
The essential diagnostic is straightforward: compute the two-sided limit. If it exists as a finite number L, the discontinuity is removable, and you redefine f(a) = L. If the one-sided limits exist but differ, you have a jump. If either one-sided limit is infinite, you have a vertical asymptote. On the AP exam, the algebraic signal for a removable discontinuity in a rational function is a 0/0 indeterminate form at the point of interest, meaning both numerator and denominator share a common factor.
Worked Example
Consider the function f(x) = (x² − 5x + 6)/(x² − 4). We want to determine whether f has a removable discontinuity at x = 2 and, if so, redefine the function to make it continuous there.
Techniques for Finding Limits at Removable Discontinuities
The 0/0 indeterminate form that signals a removable discontinuity can be resolved by several algebraic techniques. The table below compares the major methods, each suited to different function types. Being fluent in all of them gives you flexibility on the AP exam, where the function may involve polynomials, radicals, or trigonometric expressions.
| Technique | When to Use | Example |
|---|---|---|
| Factoring | Rational functions where numerator and denominator share a common polynomial factor | (x² − 9)/(x − 3) = (x + 3) for x ≠ 3 |
| Rationalization | Expressions involving square roots that produce 0/0; multiply by the conjugate | (√x − 2)/(x − 4): multiply by (√x + 2)/(√x + 2) |
| Trig identities | Trigonometric 0/0 forms; use sin(x)/x → 1 or Pythagorean identities | sin(3x)/x = 3 · [sin(3x)/(3x)] → 3 as x → 0 |
| Simplifying complex fractions | Nested fractions that produce 0/0; combine into a single fraction first | (1/x − 1/3)/(x − 3): combine numerator, then cancel |
| Piecewise redefinition | When f is defined piecewise and a parameter must be chosen to force continuity | f(x) = {x² + k, x ≤ 1; 3x, x > 1}: set 1 + k = 3 |
Connection to Advanced Topics
Removing discontinuities is not merely a stand-alone technique; it is foundational for several advanced topics in calculus. The very definition of the derivative relies on evaluating a limit that initially appears as a 0/0 indeterminate form. Understanding how to handle removable discontinuities prepares you for differentiation, L'Hôpital's Rule (covered in BC and college courses), and the analysis of functions defined by integrals.
| Concept in This Lesson | Advanced Connection |
|---|---|
| Evaluating lim(x→a) [f(x) − f(a)]/(x − a) as a 0/0 form | This is the definition of f′(a). Every derivative computation is, in essence, removing a removable discontinuity of the difference quotient. |
| Factoring and cancelling common factors | L'Hôpital's Rule (AP Calculus BC) provides an alternative when algebraic simplification is intractable: differentiate numerator and denominator separately. |
| Redefining f(a) = L to achieve continuity | In real analysis, this idea generalizes to extensions of functions. A function continuous on (a, b) can sometimes be uniquely extended to [a, b] by assigning limit values at endpoints. |
| Piecewise functions with parameters | Differential equations often require matching conditions at boundaries—an extension of ensuring piecewise-defined solutions are continuous. |
When you encounter the formal definition of the derivative in the next unit—f′(a) = lim(h→0) [f(a + h) − f(a)]/h—remember that plugging in h = 0 directly gives 0/0. You will need the same factoring, cancelling, and limit-evaluation skills you have practiced in this lesson. In that sense, mastering removable discontinuities is rehearsal for every differentiation problem to come.
Practice Problems
Lesson Summary
A removable discontinuity occurs at x = a when the two-sided limit lim(x→a) f(x) = L exists and is finite, but the function either is undefined at a or takes a value different from L. To remove the discontinuity, redefine f(a) = L. This is the only type of discontinuity that can be repaired by a single-point redefinition; jump discontinuities and infinite discontinuities (vertical asymptotes) cannot be removed.
The algebraic techniques for evaluating the limit at a removable discontinuity include factoring and cancelling common factors, rationalization (multiplying by a conjugate), and applying trigonometric identities such as sin(u)/u → 1. On the AP exam, expect to encounter these in both multiple-choice and free-response contexts, including piecewise functions where you must determine a parameter value that ensures continuity at a specified point. The concept is also the algebraic foundation for the definition of the derivative, which is itself a limit of a 0/0 difference quotient.