How do I know if a discontinuity is removable before trying to fix it?

Check whether the two-sided limit _x → a f(x) exists. If it does but doesn't match f(a) (or f(a) is undefined), it's removable. If the left and right limits differ (jump) or the function blows up (infinite), it is non-removable and cannot be patched.

What value do I assign to make the function continuous?

Always the limit: set f(a) = _x → a f(x). Compute that limit by factoring and canceling the common factor, then substitute. Don't use the value that makes the denominator zero or the raw numerator value.

Why can't I just cancel the factor and call it a day?

Canceling gives a simplified expression that equals the original everywhere except at x = a. The original still has a hole there. You must explicitly redefine f(a) equal to the limit to actually remove the discontinuity.

How is solving for a parameter like k different from finding f(a)?

It's the same idea applied in reverse. You compute the limit from the surrounding expression, then set the unknown constant equal to it so the continuity equation _x → a f(x) = f(a) holds. One unknown needs one continuity condition.

AP-CALCBC-1.13

U1.13 Removing Discontinuities

Learn to remove removable discontinuities in AP Calculus BC: find f(a) or an unknown parameter that makes a function continuous using limits.

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A voice-first session with the Crimsora tutor on U1.13 Removing Discontinuities, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

A removable discontinuity is a hole in a graph — the limit exists, but the function value is missing or misplaced. The good news? You can fix it. In this lesson you'll learn exactly what value

f(a)

must take, or what unknown constant makes a piecewise function seamless.

This skill combines everything you've practiced with algebraic limits and the definition of continuity. On the AP exam it shows up as "find the value of

k

" or "redefine

f

so it is continuous" problems, and it rewards clean factoring and a solid grip on the three-part continuity test.

What "Removing" a Discontinuity Means

A function

f

is continuous at

x = a

when three conditions hold:

f(a)

is defined,

\lim_{x \to a} f(x)

exists, and

\lim_{x \to a} f(x) = f(a)

.

A discontinuity is removable precisely when the limit exists but the third condition fails — either

f(a)

is undefined (a hole) or

f(a)

is defined but equals the wrong value (a point that jumped off the curve). Because the limit already exists, we can "remove" the gap by redefining

f(a)

to equal that limit.

The rule is simple and worth memorizing: to remove a discontinuity at

x = a

, set

f(a) = \lim_{x \to a} f(x).

Contrast this with a jump discontinuity (left and right limits differ) or an infinite discontinuity (vertical asymptote). Those are non-removable — no single value of

f(a)

can patch them, because the two-sided limit does not exist. The exam frequently checks whether you can tell the difference before you attempt a fix. If the limit doesn't exist, the correct answer is "cannot be made continuous."

The Standard Procedure

Most removable-discontinuity problems involve a rational expression where a common factor cancels. The step-by-step method:

Step	Action
1	Confirm the discontinuity type — plug in $x=a$ ; a $\frac{0}{0}$ form signals a removable hole
2	Simplify: factor and cancel the common factor causing the zero denominator
3	Evaluate the limit by substituting $a$ into the simplified expression
4	Assign $f(a)$ equal to that limit value

For example,

g(x) = \frac{x^2 - 9}{x - 3}

is undefined at

x = 3

. Factor:

\frac{(x-3)(x+3)}{x-3} = x + 3

for

x \neq 3

. So

\lim_{x \to 3} g(x) = 6

, and defining

g(3) = 6

removes the hole.

A common misconception is that canceling the factor "changes" the function everywhere — it doesn't. The simplified form only agrees with the original away from

x = a

; the hole still exists in the original until you explicitly redefine the value there.

Solving for an Unknown Parameter

The richer AP version gives a piecewise function with an unknown constant and asks you to choose it so

f

is continuous. Here you use the definition directly: force the piece's limit to match the defined value.

Suppose

f(x) = \frac{x^2 - 4}{x - 2}

for

x \neq 2

and

f(2) = k

. Continuity requires

k = \lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2}(x+2) = 4

, so

k = 4

.

When a parameter sits inside the expression, set the one-sided limits equal at the boundary. For a piecewise

f

with

f(x) = ax + 1

for

x \le 3

and

f(x) = x^2 - 2

for

x > 3

, continuity at

x = 3

demands

3a + 1 = 9 - 2 = 7

, giving

a = 2

.

The key idea: continuity is one equation. If you have one unknown, one equation solves it. If a problem asks for two unknowns (say

a

and

b

), you'll typically need two conditions — often continuity plus differentiability, which appears later in the course.

How the Exam Tests This

Removing discontinuities appears in both multiple-choice and free-response settings. Multiple-choice items often give a rational function and ask "what value should be assigned to

f(a)

" — the trap answers include the value that makes the denominator zero, or the numerator's value, rather than the true limit.

Free-response prompts frequently embed this in a continuity argument: you must state that the limit exists, compute it, and explicitly say "define

f(a)

= [value]." Graders want to see the limit computation, not just the final number.

A frequent misconception: students try to "remove" a jump or infinite discontinuity. Always verify the two-sided limit exists first. Another trap is stopping after canceling — you must evaluate the limit and state the assignment. Also watch the wording: "can this discontinuity be removed?" expects a yes/no justified by whether

\lim_{x \to a} f(x)

exists. On calculator-active sections you can confirm a hole numerically by evaluating

f

near

a

from both sides, but you still need the exact value algebraically.

Key terms

Removable discontinuity.: A discontinuity at $x=a$ where $\lim_{x \to a} f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined); it can be fixed by redefining $f(a)$ .
Continuity at a point.: $f$ is continuous at $a$ if $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the two are equal.
Indeterminate form $\frac{0}{0}$ .: The result of direct substitution that signals a possible removable discontinuity, resolvable by factoring and canceling.
Non-removable discontinuity.: A jump or infinite discontinuity where the two-sided limit does not exist, so no assignment of $f(a)$ restores continuity.
Parameter.: An unknown constant (like $k$ or $a$ ) whose value is chosen to satisfy a continuity condition.
One-sided limit matching.: Setting $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$ at a piecewise boundary to enforce continuity.

Worked example

Let

f(x) = \dfrac{x^2 - x - 6}{x - 3}

for

x \neq 3

. Determine the value that should be assigned to

f(3)

so that

f

is continuous at

x = 3

First check the discontinuity type. Substituting

x = 3

into the numerator gives

9 - 3 - 6 = 0

, and the denominator gives

0

, so we have the indeterminate form

\frac{0}{0}

. This indicates a removable discontinuity, so a fix is possible.

Factor the numerator:

x^2 - x - 6 = (x - 3)(x + 2)

.

Simplify for

x \neq 3

\dfrac{(x-3)(x+2)}{x-3} = x + 2

.

Evaluate the limit:

\lim_{x \to 3} f(x) = \lim_{x \to 3}(x + 2) = 5

.

Because the limit exists and equals

5

, continuity requires

f(3) = \lim_{x \to 3} f(x)

. Therefore assign

f(3) = 5.

With this definition, all three continuity conditions are satisfied and the hole at

x = 3

is removed.

Practice questions

The function

h(x) = \dfrac{x^2 - 25}{x - 5}

has a removable discontinuity at

x = 5

. What value should be assigned to

h(5)

to make

h

continuous there?

$0$
$5$
$10$
$25$

Answer: $10$

Factor the numerator:

x^2 - 25 = (x-5)(x+5)

. Cancel to get

x + 5

for

x \neq 5

. Then

\lim_{x \to 5} h(x) = 5 + 5 = 10

. Continuity requires

h(5) = 10

. The trap answer

0

comes from plugging into the denominator, and

25

from misreading the numerator.

Consider

f(x) = \dfrac{\sin(2x)}{x}

for

x \neq 0

. Explain whether the discontinuity at

x = 0

is removable, and if so, state the value

f(0)

should take.

Answer: The discontinuity is removable; define $f(0) = 2$ .

Direct substitution gives

\frac{0}{0}

, so investigate the limit. Rewrite

\frac{\sin(2x)}{x} = 2 \cdot \frac{\sin(2x)}{2x}

. Using

\lim_{u \to 0}\frac{\sin u}{u} = 1

with

u = 2x

, the limit is

2 \cdot 1 = 2

. Since the two-sided limit exists and equals

2

, the discontinuity is removable and continuity requires

f(0) = 2

Find the value of the constant

k

that makes

f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}

continuous at

x = 3

Answer: $k = 4$

Factor:

x^2 - 2x - 3 = (x-3)(x+1)

, which cancels to

x + 1

for

x \neq 3

. So

\lim_{x \to 3} f(x) = 3 + 1 = 4

. Continuity at

x = 3

requires

k = \lim_{x \to 3} f(x) = 4

FAQ

How do I know if a discontinuity is removable before trying to fix it?: Check whether the two-sided limit $\lim_{x \to a} f(x)$ exists. If it does but doesn't match $f(a)$ (or $f(a)$ is undefined), it's removable. If the left and right limits differ (jump) or the function blows up (infinite), it is non-removable and cannot be patched.
What value do I assign to make the function continuous?: Always the limit: set $f(a) = \lim_{x \to a} f(x)$ . Compute that limit by factoring and canceling the common factor, then substitute. Don't use the value that makes the denominator zero or the raw numerator value.
Why can't I just cancel the factor and call it a day?: Canceling gives a simplified expression that equals the original everywhere except at $x = a$ . The original still has a hole there. You must explicitly redefine $f(a)$ equal to the limit to actually remove the discontinuity.
How is solving for a parameter like $k$ different from finding $f(a)$ ?: It's the same idea applied in reverse. You compute the limit from the surrounding expression, then set the unknown constant equal to it so the continuity equation $\lim_{x \to a} f(x) = f(a)$ holds. One unknown needs one continuity condition.

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