1.13 Removing Discontinuities
In this guide, we’ll be diving into the fascinating world of discontinuities. Ever looked at a graph and noticed it suddenly jumps or has holes? These are discontinuities, and sometimes, we can remove them. Let's explore how we can smooth out these mathematical speed bumps and make our functions continuous.
⭕ What are Discontinuities?
Before we patch things up, let's first understand what we're dealing with. Discontinuities occur where a function "breaks." There are three types: removable, jump, and infinite. Today, we are going to focus on removable discontinuities.
💡 Jump back to 1.12 if you need a refresher on confirming continuity over an interval!
📍Removable Discontinuities
Removable discontinuities are also sometimes called “holes” in a graph. They exist when a function is undefined at a point, but its limit exists. We can fill this hole to make the graph continuous. Here’s an example:
We can see in this graph, the limit as approaches exists, it’s just undefined at that point. But, if we color in the circle (in other words, remove the discontinuity), the graph will be continuous over the interval in the image.
🌀 Filling the Gap
To remove a discontinuity, we redefine the function's value at that point to equal the limit of the function as x approaches that point. For example:
To remove this discontinuity, we can factor the denominator and cancel like terms, like so:
This function no longer has a discontinuity at , instead, it is defined as .
✏️ Practice Filling the Gap
Consider the function defined as follows:
where is a constant. Determine the value of that would make continuous at .
Here's the solution! ⬇️
First, we need to factor our numerator. This will allow us to see whether we can cancel out the denominator, which is what is causing our discontinuity.
We see that there is an term in our numerator, so we can cancel it with the in the denominator!
For ,
📈 Piecewise Functions
For piecewise functions, we can ensure continuity by checking the right and left limit of the function, and assuring that the value is the same as the one defined at the point.
✔️ Ensuring Continuity
Consider defined by two pieces of a function on either side of :
- On the left side, approaches as approaches .
- On the right side, approaches as approaches .
For to be continuous at , we need .
For example, this image depicts the continuous, piecewise function
We can see the the limit of as it approaches the point from the left is equal to , and similarly, from the right, it is also . Finally, is defined at as . Thus, our function is continuous.
✏️ Practice Ensuring Continuity
Consider the function,
What must be set to for the function to be continuous?
To solve this problem, we will use the first part of the piecewise function to solve for at . First, let’s factor and cancel some terms:
Now, let’s plug in :
Finally, we set this equal to $a$ and solve:
📷 Visualizing Continuity
Graphs are an excellent way to see continuity (or the lack thereof). Use graphing tools to visualize the function and identify discontinuities. A continuous graph can be drawn without lifting your pencil!
✏️ Practice Visualizing Continuity
Graph the function over the interval . Is it continuous? Can you make it continuous?
The graph of this function looks like so:
You can see that there is a discontinuity at . But, we can remove it by factoring!
Plugging in , we find that at , making our function continuous.
⭐ Closing
📚 AP Calc is practice-driven! Attempt more problems, especially from past AP exams, to strengthen your understanding. Always check for continuity and practice "patching up" those functions. Great work! 👏
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| continuous | A function that has no breaks, jumps, or holes in its graph over a given interval. |
| discontinuity | A point where a function is not continuous, often due to a break, jump, or hole in the graph. |
| limit | The value that a function approaches as the input approaches some value, which may or may not equal the function's value at that point. |
| piecewise-defined function | A function defined by different expressions over different intervals or regions of its domain. |
| removable discontinuity | A discontinuity that can be eliminated by defining or redefining the function value at that point to equal the limit. |
Frequently Asked Questions
How do I remove a discontinuity from a function?
A removable discontinuity (a “hole”) can be fixed only when the limit exists at that point. Quick steps: 1. Identify the point a where f is undefined or piecewise changes. 2. Compute the two one-sided limits: lim_{x→a^-} f(x) and lim_{x→a^+} f(x). If they’re equal, call that common value L. (AP expects you check one-sided limits for piecewise boundaries—EK LIM-2.C.1 and EK LIM-2.C.2.) 3. If the limit L exists, redefine f(a) = L to remove the discontinuity. If the one-sided limits differ or limit is infinite, it’s not removable. 4. For algebraic rational functions, factor and cancel the common (x−a) factor, then evaluate the simplified expression at a to find L. Example: f(x) = (x^2−1)/(x−1). Factor to (x+1), so lim_{x→1} f(x)=2. Redefine f(1)=2 to remove the hole. For piecewise functions, solve for parameters so left-hand expression at the boundary equals the right-hand value and the function value there. More practice: see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and unit review (https://library.fiveable.me/ap-calculus/unit-1).
What's the difference between removable and non-removable discontinuities?
A removable discontinuity (a “hole”) is a point where the two one-sided limits agree (so the overall limit exists) but the function’s value is either undefined or not equal to that limit. You can “remove” it by redefining the function at that x to equal the limit—common with rational functions after factor cancellation. This is exactly EK LIM-2.C.1: if the limit exists at the discontinuity, you can redefine the value to make the function continuous. A non-removable discontinuity is one you can’t fix just by redefining the function at that single x. Examples: a vertical asymptote (limit is infinite) or a jump/step where left- and right-hand limits are different (so the limit does not exist). For piecewise functions, continuity at a boundary requires left value = right value = function value (EK LIM-2.C.2); if that fails and the one-sided limits differ, it’s non-removable. For extra practice and examples, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and lots of problems (https://library.fiveable.me/practice/ap-calculus).
When can I actually fix a discontinuity and when can't I?
You can “fix” a discontinuity only when the limit exists at that x—meaning the left-hand and right-hand limits are equal. If lim f(x) as x→a exists but f(a) is missing or different, you can redefine f(a) to equal that limit and remove the hole (a removable discontinuity; see EK LIM-2.C.1). Typical algebraic case: a factor cancels in a rational function so you get a hole you can fill by taking the simplified value. You can’t fix it when one-sided limits differ (a jump) or when the function blows up (vertical asymptote/infinite limit)—there’s no single finite value to assign that makes the function continuous. For piecewise functions, continuity at a boundary needs left limit = right limit = function value at the boundary (EK LIM-2.C.2). For more examples and AP-style practice, check the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and unit review (https://library.fiveable.me/ap-calculus/unit-1). Practice problems are at (https://library.fiveable.me/practice/ap-calculus).
How do I solve for the parameter that makes a piecewise function continuous?
To make a piecewise function continuous at a boundary x = a, you need the left-hand limit, right-hand limit, and the function’s value at a to all match (CED EK LIM-2.C.2). Practically: 1. Compute lim(x→a−) f(x) using the expression for x
I don't understand how to make a function continuous at a point - can someone explain step by step?
Short version: continuity at a point means limit exists there and the function’s value equals that limit. Step-by-step: 1. Identify the point a. Check one-sided limits: lim x→a− f(x) and lim x→a+ f(x). If they’re not equal, the limit doesn’t exist → not removable. (CED: one-sided limits, limit exists.) 2. If one-sided limits match, compute L = lim x→a f(x). If f(a) is undefined or f(a) ≠ L, you can remove the discontinuity by redefining f(a) = L (removable hole / removable discontinuity). 3. For algebraic functions (like rationals), simplify: factor and cancel common (x−a) terms, then plug in a to find L. 4. For piecewise functions, set the left-expression and right-expression equal at the boundary and solve for any parameter(s) so both sides (and f(a) if given) equal the same value (CED: continuity at a boundary). 5. Always mention one-sided limits on the AP—they assess LIM-2.C and piecewise continuity. For worked notes and examples see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD). For more review and practice, check the Unit 1 overview (https://library.fiveable.me/ap-calculus/unit-1) and practice problems (https://library.fiveable.me/practice/ap-calculus).
What's the formula for removing discontinuities in rational functions?
If a rational function has a removable discontinuity (a “hole”) at x = a, the fix is: redefine the function value at a to equal the limit as x → a. Practically: 1. Factor numerator and denominator and cancel any common factor (x − a). 2. Evaluate the simplified expression at x = a to get L = lim_{x→a} f(x). 3. Redefine f(a) = L. That removes the hole if the one-sided limits agree (i.e., the limit exists). In symbols: if f(x) = p(x)/q(x) and p(x) = (x−a)·r(x), q(x) = (x−a)·s(x) with s(a) ≠ 0, then for x ≠ a f(x) = r(x)/s(x) and lim_{x→a} f(x) = r(a)/s(a). Define f(a) := r(a)/s(a) to make f continuous at a (EK LIM-2.C.1, factor cancellation). For piecewise functions, set the boundary value equal to the common one-sided limit (EK LIM-2.C.2). For more examples and practice, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and Unit 1 overview (https://library.fiveable.me/ap-calculus/unit-1).
How do I know if a discontinuity is removable just by looking at the graph?
Look for a hole (an open circle) in the curve where the left- and right-hand behavior match. If the graph approaches the same y-value from both sides but either there's an open dot (no value) or a filled dot at a different y, that point is a removable discontinuity—you can redefine f(a) to equal the common limit. If the left and right limits are different (the two sides go to different y-values) the discontinuity is nonremovable. For piecewise functions, check the boundary value: the two side-expressions and f(a) must agree for continuity (EK LIM-2.C.2). In AP language: a discontinuity is removable exactly when the limit exists at that x (EK LIM-2.C.1). For practice and quick examples, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and try problems at (https://library.fiveable.me/practice/ap-calculus).
When do I use limits to fix discontinuities vs just plugging in values?
Use limits whenever the function’s value at the point is undefined or doesn’t match the behavior of nearby x-values. Key CED ideas: - If f(a) is defined and lim_{x→a} f(x) exists and equals f(a), the point is continuous (EK LIM-2.C.1). If f(a) ≠ lim, you can “remove” the hole by redefining f(a) to equal the limit. - If you get an indeterminate form like 0/0, algebraically simplify (factor/cancel) and then plug the simplified expression into x = a to find the limit—that limit is the value you’d assign to remove a removable discontinuity. - If one-sided limits differ (jump) or either side is infinite, the two-sided limit doesn’t exist so you can’t fix continuity by one value. - For piecewise functions, require left-hand limit = right-hand limit = function value at the boundary (EK LIM-2.C.2). Quick checklist: Is f(a) defined? Is lim_{x→a} f(x) finite? Do one-sided limits match? If yes, plug that limit in; if no, you must use limits to analyze and you can’t always “fix” the discontinuity. More on examples and practice in the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD). For broad review: (https://library.fiveable.me/ap-calculus/unit-1). For extra practice problems: (https://library.fiveable.me/practice/ap-calculus).
How do I make a piecewise function continuous at the boundary points?
Focus on each boundary point c. For a piecewise function you need three things at x = c:
1) left-hand limit exists: lim x→c− f(x),
2) right-hand limit exists: lim x→c+ f(x),
3) those two one-sided limits are equal and equal to the function value f(c). So if your pieces are g(x) for x
What does it mean that the limit exists at a discontinuity but the function value doesn't?
If the limit exists at a discontinuity but the function value doesn’t, it means the function’s behavior approaching that x-value settles to a single number L (left-hand and right-hand limits equal L), but f(a) is either undefined or equals some other number. Graphically that’s a removable discontinuity (a “hole”)—the curve approaches a point but there’s no dot (or the dot is in the wrong place). By EK LIM-2.C.1 you can “remove” the discontinuity by redefining f(a) = L so the function becomes continuous at a. Example: f(x) = (x^2 − 1)/(x − 1) simplifies to x + 1 for x ≠ 1, so lim x→1 f(x) = 2, but f(1) is undefined. Redefine f(1)=2 to remove the hole. For AP-style piecewise problems, remember EK LIM-2.C.2: both one-sided expressions and the function value must match at the boundary (see the Topic 1.13 study guide: https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD). For extra practice try the AP practice question bank (https://library.fiveable.me/practice/ap-calculus).
I'm confused about redefining function values - how do I know what to redefine it as?
You redefine the value at a point so the function equals the limit as x approaches that point—but only when that two-sided limit exists (EK LIM-2.C.1). Practically: - Check the limit. If lim x→a f(x) exists, define f(a) = that limit to “fill the hole.” - For piecewise functions, make left- and right-hand expressions match at the boundary and set f(boundary) equal to that common value (EK LIM-2.C.2). Common algebra steps: simplify the expression (factor and cancel removable factors), then plug in x = a. Example: f(x) = (x^2 − 1)/(x − 1) is undefined at 1, but simplifies to x + 1, so lim x→1 f(x) = 2—redefine f(1)=2. On the AP exam you’ll be asked to solve for parameter values or boundary values that make a function continuous; show the one-sided limits when relevant and state the value you redefine to (the limit). For more practice and worked examples, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and the Unit 1 overview (https://library.fiveable.me/ap-calculus/unit-1). For extra problems, try the practice bank (https://library.fiveable.me/practice/ap-calculus).
How do I solve problems where I need to find the value of k that makes the function continuous?
You want k so the function has no removable hole at x = a—so make f(a) equal the limit as x → a. Steps: 1. Check the two one-sided limits (if piecewise) or the two-sided limit. For continuity you need lim x→a− f(x) = lim x→a+ f(x) = L and f(a) = L (EK LIM-2.C.1 & LIM-2.C.2). 2. If f is algebraic/rational with a factor (x − a) causing 0/0, factor and cancel to find the simplified expression g(x). Then L = g(a). Set k = L. 3. If the function is piecewise, evaluate each side at the boundary and solve for k so left value = right value = k. 4. If one-sided limits differ or limit doesn’t exist, no value of k can make it continuous. Example outline: f(x) = { (x^2−1)/(x−1) for x≠1; k for x=1 } → simplify to x+1, so L = 2, choose k = 2. For guided examples and AP-aligned practice, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and more practice problems (https://library.fiveable.me/practice/ap-calculus).
What's the step-by-step process for checking if a piecewise function is continuous everywhere?
Step-by-step check for continuity of a piecewise function: 1. Identify pieces and their domains. List the boundary points where the formula changes (and any endpoints or domain holes). 2. For every x inside a piece (not on a boundary), check the defining expression is continuous there (polynomials, exponentials, trig, rational without zero denom.—use algebra). If each piece’s rule is continuous on its open interval, those interior points are fine. 3. At each boundary point a between pieces: - Compute the left-hand limit L = lim_{x→a^-} f(x). - Compute the right-hand limit R = lim_{x→a^+} f(x). - Evaluate f(a) if it's defined. - For continuity at a you need L = R = f(a). If L = R but f(a) ≠ that common value (or f(a) is undefined), it’s a removable discontinuity—redefine f(a) = L to make it continuous (EK LIM-2.C.1/2). 4. Check domain endpoints similarly using one-sided limits. 5. If any one-sided limits differ, the point is a jump/nonremovable discontinuity. On the AP exam, be explicit: show the one-sided limits and compare with f(a) (CED LIM-2.C language). For practice and worked examples, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and more problems (https://library.fiveable.me/practice/ap-calculus).
Why can't I remove jump discontinuities but I can remove holes?
A removable discontinuity (a hole) happens when the two one-sided limits are equal—the limit exists—but the function’s value at that x is missing or different. Because EK LIM-2.C.1 says you can redefine f(a) to equal the limit, you “remove” the hole and make the function continuous there. A jump discontinuity happens when the left-hand and right-hand limits are different (the one-sided limits don’t match). The two-sided limit does not exist, so there’s no single number you can assign to f(a) that matches both sides. That’s why a jump can’t be removed by redefining the function at a single point. For piecewise boundaries (EK LIM-2.C.2) you need the left expression, right expression, and the value at the point all equal to be continuous—if the left and right values disagree, continuity is impossible at that point. For more practice and examples (including graphs and algebraic cancellation for holes), check the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and try problems on Fiveable’s AP Calc practice page (https://library.fiveable.me/practice/ap-calculus).
How do I use the three conditions for continuity to remove discontinuities?
The three conditions for continuity at a point a are: (1) f(a) is defined, (2) the limit lim_{x→a} f(x) exists (i.e., left-hand and right-hand limits are equal), and (3) f(a) = lim_{x→a} f(x). To remove a removable discontinuity, use them in this order: 1. Check one-sided limits. Compute lim_{x→a^-} f(x) and lim_{x→a^+} f(x). If they aren’t equal the discontinuity isn’t removable. 2. If they are equal, that common value L = lim_{x→a} f(x) exists. For algebraic functions, simplify (factor/cancel) to find L. 3. Redefine or choose the parameter so f(a) = L. For piecewise functions, set the expression values from each side equal at the boundary and solve for the parameter; then set the function value at a to that common value. Example: f(x) = (x^2−1)/(x−1) for x≠1. Simplify to x+1, so limit as x→1 is 2. Define f(1)=2 to remove the hole. This is exactly what the CED expects for LIM-2.C tasks. For worked examples and practice, see the Topic 1.13 study guide (https://library.fiveable.me/ap-calculus/unit-1/removing-discontinuities/study-guide/InY5nKkX7B44HNb3T0fD) and Unit 1 resources (https://library.fiveable.me/ap-calculus/unit-1). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).