Removing Discontinuities - Calculus 1

SECTION 1

Historical Context & Motivation

For centuries, mathematicians wrestled with the idea of smooth, unbroken curves. When Isaac Newton and Gottfried Leibniz developed calculus in the late 1600s, they relied on an intuitive understanding of functions flowing continuously from one value to the next. However, they quickly encountered situations where a function's formula broke down at a single point — producing a discontinuity — even though the surrounding values behaved perfectly well. The concept of removing discontinuities emerged as mathematicians realized that many of these breaks were artificial, caused by the algebra of a formula rather than any genuine gap in the curve.

1684 • Leibniz Publishes His Calculus

Gottfried Leibniz published the first paper on differential calculus, using the idea that curves change smoothly. Functions with breaks or holes didn't fit neatly into this framework.

1821 • Cauchy Defines Continuity

Augustin-Louis Cauchy gave the first rigorous definition of continuity using limits, allowing mathematicians to precisely identify where a function was — and was not — continuous.

1861 • Weierstrass Formalizes Limits

Karl Weierstrass introduced the epsilon-delta definition of limits, making it possible to prove rigorously whether a discontinuity could be "removed" by redefining a function at a single point.

1900s • Modern Textbook Treatment

Removable discontinuities became a standard topic in introductory calculus courses, serving as a bridge between the algebraic manipulation of functions and the analytical power of limits.

The central question this concept addresses is straightforward: if a function has a "hole" at a particular x-value — meaning the formula is undefined or assigned the wrong value there — can we fill in that hole so the function becomes continuous? As you'll see, the answer depends entirely on whether the limit exists at that point.

SECTION 2

Core Principles & Definitions

Before you can remove a discontinuity, you need to understand what makes a function continuous in the first place. A function f(x) is continuous at x = a if three conditions are all satisfied: the function is defined at a, the limit of f(x) as x approaches a exists, and the limit equals the function value f(a). When any one of these conditions fails, the function has a discontinuity at that point. The type of failure determines whether the discontinuity can be fixed.

Removable Discontinuity

The limit of f(x) as x → a exists, but either f(a) is undefined or f(a) does not equal the limit. This creates a single "hole" in the graph that can be patched.

Jump Discontinuity

The left-hand limit and right-hand limit both exist but are not equal. The graph "jumps" from one value to another. This type cannot be removed.

Infinite Discontinuity

The function approaches positive or negative infinity near x = a, producing a vertical asymptote. The limit does not exist as a finite number, so this cannot be removed.

The Removal Process

To remove a discontinuity, redefine f(a) so that it equals lim(x→a) f(x). This fills in the hole and makes the function continuous at x = a.

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KEY TAKEAWAY

Think of a removable discontinuity like a single missing tile on a sidewalk. The path is complete on both sides, and you can clearly see what size and shape the missing tile should be. You just need to place the right tile — the value of the limit — into the gap. A jump or infinite discontinuity, on the other hand, is like a broken bridge: the two sides don't line up, and no single tile can fix it.

SECTION 3

Visualizing Removable Discontinuities

The best way to understand a removable discontinuity is to see one. Consider the function f(x) = (x² − 4)/(x − 2). If you try to evaluate f(2), you get 0/0 — the function is undefined there. But the graph of this function looks almost identical to the line y = x + 2, except for a single open circle (a "hole") at the point (2, 4). The diagram below shows this function with its discontinuity and the repaired version side by side.

Left: the original function f(x) = (x² − 4)/(x − 2) has an open circle (hole) at x = 2 because the formula produces 0/0. Right: the repaired function g(x) = x + 2 fills the hole with a solid dot at (2, 4), making the function continuous.

In the left graph, the open circle at (2, 4) tells you that f(2) is not defined — the function has a removable discontinuity there. Notice, though, that as x approaches 2 from either side, the y-values approach 4. The limit exists and equals 4. In the right graph, we define g(x) = x + 2 for all x, which automatically assigns g(2) = 4. The solid dot confirms that the hole is patched and the function is now continuous at x = 2.

SECTION 4

Mathematical Framework

The formal process of removing a discontinuity rests on the definition of continuity. A function f is continuous at x = a when three conditions hold simultaneously. If the first two conditions are met but the third fails — or the first condition fails altogether — you may have a removable discontinuity.

CONTINUITY AT A POINT

f is continuous at x = a ⟺ (1) f(a) is defined, (2) lim(x→a) f(x) exists, (3) lim(x→a) f(x) = f(a)

All three conditions must be true. If condition (2) holds but (1) or (3) fails, the discontinuity is removable.

REMOVABLE DISCONTINUITY CONDITION

lim(x→a) f(x) = L exists as a finite number, but f(a) ≠ L or f(a) is undefined

Here L is a real number. The key insight is that the limit must exist — both the left-hand and right-hand limits must agree.

THE FIX: REDEFINE THE FUNCTION

g(x) = { f(x), if x ≠ a; L, if x = a } where L = lim(x→a) f(x)

The new function g is identical to f everywhere except at x = a, where it takes the value of the limit. This makes g continuous at a.

In practice, you usually find the limit by simplifying the algebraic expression. Common techniques include factoring and canceling common factors, rationalizing a numerator or denominator that contains a radical, and simplifying complex fractions. Once you simplify and evaluate the limit, you know the value that fills the hole.

SECTION 5

Types of Removable Discontinuities & Algebraic Techniques

Removable discontinuities arise from different algebraic structures. Recognizing the structure helps you choose the right simplification strategy. The most common scenarios are described below, along with a visual flowchart to guide your approach.

Scenario	Example	Technique
Common linear factor in numerator and denominator	(x² − 9)/(x − 3)	Factor both, cancel (x − 3), then evaluate
Radical expression producing 0/0	(√(x + 1) − 2)/(x − 3)	Multiply by the conjugate of the radical expression
Complex fraction yielding 0/0	(1/x − 1/3)/(x − 3)	Combine the sub-fractions, then simplify and cancel
Piecewise function with wrong value at a point	f(x) = x + 1 for x ≠ 2, f(2) = 5	Compute the limit, redefine f(2) to equal it

Common scenarios that produce removable discontinuities

Start at the top by attempting direct substitution. If you get the indeterminate form 0/0, follow the right-hand path to simplify, cancel, and evaluate. If direct substitution gives a non-zero number over zero (suggesting ±∞), the discontinuity is not removable.

The 0/0 result is your signal — it means the numerator and denominator share a common factor that vanishes at x = a. Once you cancel that factor, the expression simplifies to something you can evaluate directly. The result of that evaluation is the limit, and therefore the value you assign to fill the hole.

SECTION 6

Worked Example: Removing a Discontinuity

Let's walk through a complete example. We'll find and remove the discontinuity in the function f(x) = (x² − 5x + 6)/(x − 2).

Remove the discontinuity in f(x) = (x² − 5x + 6)/(x − 2)

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Step 1 — Attempt Direct Substitution

Substitute x = 2 into the function: f(2) = (2² − 5(2) + 6)/(2 − 2) = (4 − 10 + 6)/(0) = 0/0. Since we get the indeterminate form 0/0, the function is undefined at x = 2, but a removable discontinuity may exist.

f(2) = 0/0 → indeterminate form detected
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Step 2 — Factor the Numerator

Factor x² − 5x + 6. We look for two numbers that multiply to 6 and add to −5. Those numbers are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3).

x² − 5x + 6 = (x − 2)(x − 3)
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Step 3 — Cancel the Common Factor

Rewrite the function: f(x) = (x − 2)(x − 3)/(x − 2). For all x ≠ 2, we can cancel the common factor (x − 2) from numerator and denominator, leaving f(x) = x − 3 for x ≠ 2.

f(x) = x − 3 (for x ≠ 2)
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Step 4 — Evaluate the Limit

Now evaluate the simplified expression at x = 2: lim(x→2) f(x) = 2 − 3 = −1. The limit exists and equals −1.

lim(x→2) f(x) = −1
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Step 5 — Redefine the Function

Define a new continuous function g(x) that equals f(x) everywhere except at the discontinuity: g(x) = x − 3 for all x. Equivalently, g(x) = f(x) when x ≠ 2, and g(2) = −1. The discontinuity has been removed.

g(x) = x − 3 is continuous everywhere, including at x = 2

SECTION 7

Removable vs. Non-Removable Discontinuities

Not every discontinuity can be fixed. Understanding the differences between the three main types of discontinuity helps you quickly identify which ones are removable and which are not. The table below compares their key features.

Feature	Removable	Jump	Infinite
Graph appearance	A single hole (open circle)	Graph "jumps" between two y-values	Vertical asymptote; graph shoots to ±∞
Limit behavior	lim(x→a) f(x) = L (exists, finite)	Left and right limits exist but differ	At least one side → ±∞
Direct substitution	Gives 0/0	Function is typically piecewise	Gives nonzero/0
Can it be fixed?	Yes — redefine f(a) = L	No — the two sides disagree	No — the function blows up
Common example	(x² − 1)/(x − 1) at x = 1	Floor function ⌊x⌋ at integer values	1/(x − 3) at x = 3

Comparison of the three types of discontinuity

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KEY TAKEAWAY

The magic test is the limit. If the limit at the point of discontinuity exists as a single finite number, the discontinuity is removable — like editing a typo in a sentence. If the limit doesn't exist (because the two sides disagree or the function explodes toward infinity), the break is structural and cannot be patched with a simple redefinition.

SECTION 8

Connection to Differentiability & Advanced Topics

Removing discontinuities is not just a standalone algebraic trick — it connects to some of the most important ideas you'll encounter as you move deeper into calculus. In particular, the concept plays a critical role when you study differentiability and the definition of the derivative. The derivative itself is defined as a limit that often starts as an indeterminate 0/0 form, and you "remove the discontinuity" in the difference quotient to find the slope of the tangent line.

Concept	How Removing Discontinuities Connects
Definition of the Derivative	The limit lim(h→0) [f(a+h) − f(a)]/h is a 0/0 form. You simplify and cancel h, then evaluate — the same process as removing a discontinuity.
Continuity Requirement	A function must be continuous at a point to be differentiable there. Removing a discontinuity is sometimes the first step before checking differentiability.
L'Hôpital's Rule	This advanced technique evaluates limits of 0/0 or ∞/∞ forms by differentiating numerator and denominator. It's another tool for finding the value that removes a discontinuity.
Piecewise & Real-World Models	Engineers and scientists frequently define piecewise functions for physical models. Ensuring continuity at the boundaries between pieces often requires removing discontinuities.

How removing discontinuities connects to advanced calculus topics

As you progress into Calculus 2 and beyond, you'll see that the skill of recognizing and resolving indeterminate forms is foundational. Mastering it now with these simpler rational functions will make the transition to derivatives, integrals, and series much smoother.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the function f(x) = (x² − 1)/(x − 1) has a removable discontinuity at x = 1, even though the graph looks like a straight line everywhere else. What distinguishes this from a jump discontinuity?

Show answer

When you substitute x = 1 into f(x), both the numerator and denominator equal zero, giving the indeterminate form 0/0. This means f(1) is undefined — there's a hole in the graph. However, for every x ≠ 1, the function simplifies to (x − 1)(x + 1)/(x − 1) = x + 1, which is a continuous line. The left-hand and right-hand limits both approach 2, so the two-sided limit exists and equals 2. Because the limit exists as a single finite value, we can fill the hole by defining f(1) = 2, making the discontinuity removable. A jump discontinuity, by contrast, occurs when the left and right limits exist but are different values, so no single value can bridge the gap.

Question 2: BASIC CALCULATION

Find the value of the constant c that removes the discontinuity in the function f(x) = (x² − 9)/(x − 3) when x ≠ 3, and f(3) = c.

Show answer

Factor the numerator: x² − 9 = (x − 3)(x + 3). For x ≠ 3, the function simplifies to f(x) = (x − 3)(x + 3)/(x − 3) = x + 3. Now evaluate the limit: lim(x→3)(x + 3) = 3 + 3 = 6. To remove the discontinuity, we need f(3) to equal the limit, so c = 6. With this value, f(x) = x + 3 for all x, which is continuous everywhere.

Question 3: INTERMEDIATE

Determine whether the function f(x) = (x² − x − 6)/(x² − 4) has a removable discontinuity, a non-removable discontinuity, or both. Find the x-values of each and remove any removable discontinuity.

Show answer

Factor numerator and denominator: x² − x − 6 = (x − 3)(x + 2) and x² − 4 = (x − 2)(x + 2). The function is undefined at x = 2 and x = −2. At x = −2, the factor (x + 2) cancels: f(x) = (x − 3)/(x − 2) for x ≠ −2. The limit as x → −2 is (−2 − 3)/(−2 − 2) = (−5)/(−4) = 5/4. Since the limit exists and is finite, x = −2 is a removable discontinuity; redefine f(−2) = 5/4. At x = 2, the simplified form (x − 3)/(x − 2) has a nonzero numerator (2 − 3 = −1) over zero in the denominator, so the function approaches ±∞. This is an infinite (non-removable) discontinuity, producing a vertical asymptote at x = 2.

Question 4: APPLIED

A company models its profit per unit (in dollars) as P(x) = (x² − 100)/(x − 10), where x is the number of hundreds of units produced. The formula breaks down at x = 10 (i.e., 1000 units). Find the value the profit per unit approaches at this production level, and write a corrected function that is continuous at x = 10.

Show answer

Factor the numerator: x² − 100 = (x − 10)(x + 10). For x ≠ 10, P(x) = (x − 10)(x + 10)/(x − 10) = x + 10. The limit as x → 10 is 10 + 10 = 20 dollars per unit. This means that as production approaches 1000 units, the profit per unit approaches $20. To create a continuous model, define the corrected function: P(x) = x + 10 for all x, or equivalently, keep the original formula for x ≠ 10 and set P(10) = 20. The discontinuity in the original model was an artifact of the algebraic representation, not a real business phenomenon.

Question 5: CRITICAL THINKING

Consider f(x) = (√(x + 5) − 3)/(x − 4). Show that f has a removable discontinuity at x = 4 by finding the limit using rationalization. Then explain why factoring alone would not work for this problem.

Show answer

Direct substitution gives f(4) = (√9 − 3)/(4 − 4) = (3 − 3)/0 = 0/0, confirming an indeterminate form. To resolve it, multiply the numerator and denominator by the conjugate of the numerator: f(x) = [(√(x+5) − 3)(√(x+5) + 3)] / [(x − 4)(√(x+5) + 3)] = (x + 5 − 9) / [(x − 4)(√(x+5) + 3)] = (x − 4) / [(x − 4)(√(x+5) + 3)]. Cancel (x − 4): f(x) = 1/(√(x+5) + 3) for x ≠ 4. Now evaluate: lim(x→4) 1/(√(9) + 3) = 1/(3 + 3) = 1/6. The removable discontinuity is fixed by defining f(4) = 1/6. Factoring alone wouldn't work because the numerator contains an irrational expression (a square root), which cannot be factored into polynomial factors that cancel with the denominator. Rationalization transforms the radical into a polynomial difference, revealing the hidden common factor.

SUMMARY

Lesson Summary

Summary

A removable discontinuity occurs when a function is undefined or has the wrong value at a point x = a, even though the limit as x approaches a exists as a finite number L. The hallmark of a removable discontinuity is the indeterminate form 0/0 when you attempt direct substitution. To find L, you simplify the expression using techniques such as factoring and canceling, rationalizing, or combining complex fractions, then evaluate the simplified expression at x = a.

Once L is found, the discontinuity is removed by redefining f(a) = L. This makes the function satisfy all three conditions for continuity at that point: f(a) is defined, the limit exists, and f(a) equals the limit. Remember that jump and infinite discontinuities cannot be removed because their limits either don't exist or are not finite. Mastering this process builds the foundation for understanding derivatives, since the derivative formula itself is a limit of a 0/0 indeterminate form that must be resolved in the same way.