Can I use L'Hopital's Rule to find limits at a removable discontinuity?

Yes, L'Hopital's Rule applies whenever direct substitution yields \frac{0}{0} or \frac{\infty}{\infty}, which is exactly the indeterminate form that appears at removable discontinuities. However, for polynomial and rational functions, factoring and canceling is usually faster and avoids the need to compute derivatives.

Is a removable discontinuity the same as a hole in the graph?

Yes; a removable discontinuity appears as a hole (or open circle) on the graph at the point (a, L) where L is the limit value. The function approaches this point from both sides but never actually reaches it because f(a) is undefined or does not equal L.

How do I know whether a limit is removable or non-existent?

Factor both numerator and denominator; if they share a common factor that cancels, the discontinuity is removable and the limit exists. If no common factor cancels and you still get an indeterminate form, check the left and right limits separately to determine whether the limit truly exists.

What is the difference between a removable discontinuity and a jump discontinuity?

At a removable discontinuity, the limit exists but the function value is missing or incorrect; you can 'fix' this by redefining f(a). At a jump discontinuity, the left and right limits both exist but differ from each other — no single-point redefinition can make the function continuous because the function jumps.

Removable Discontinuity

A removable discontinuity is a hole in a graph where the limit exists but the function value differs; it can be fixed by redefining that point.

\lim_{x \to a} f(x) \text{ exists but } \neq f(a)

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What each symbol means

What Removable Discontinuity takes

Symbol	Meaning
$f$	A function representing a rule that maps inputs to outputs. The limit as $x$ approaches $a$ might exist even if $f(a)$ is undefined; mistaking a function for a number will prevent you from computing limits.
$x$	The input variable that approaches the value $a$ ; no particular units apply. If you treat $x$ as a fixed constant equal to $a$ instead of a variable approaching $a$ , you will incorrectly conclude the limit does not exist.
$a$	The fixed $x$ -value where we check for a discontinuity; no units apply. If you misread $a$ as the output value instead of the input value, you will confuse what the limit approaches with what you are measuring.

When to use it

When direct substitution gives an indeterminate form like $\frac{0}{0}$ in a rational function, suspect a removable discontinuity and factor to find the limit.

Level

Usually taught in: Calculus I

Worked examples

1. Find the removable discontinuity of a rational function

Problem

Find the removable discontinuity of

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

and state the value of

\lim_{x \to 2} f(x)

$x^2 - 4 = (x - 2)(x + 2)$
Factor the numerator as a difference of squares using $a^2 - b^2 = (a - b)(a + b)$ with $a = x$ and $b = 2$ .
$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \text{ for } x \neq 2$
Cancel the common factor $(x - 2)$ from numerator and denominator; this is valid for $x \neq 2$ since we cannot divide by zero.
$\lim_{x \to 2} (x + 2) = 2 + 2 = 4$
Evaluate the simplified expression at $x = 2$ to find the limit; the original function is undefined at this point, but the limit exists.

Answer: $\lim_{x \to 2} f(x) = 4$

The original function is undefined at $x = 2$ because the denominator is zero. Factoring reveals that $(x - 2)$ is a common factor of both numerator and denominator. After canceling, the function simplifies to $x + 2$ , which has a well-defined limit of 4 as $x \to 2$ . This is a removable discontinuity because we could 'fill in' the hole by redefining $f(2) = 4$ .

2. Identify a removable discontinuity with negative factors

Problem

Find the removable discontinuity of

g(x) = \frac{x^2 - 2x - 3}{x + 1}

and find

\lim_{x \to -1} g(x)

$x^2 - 2x - 3 = (x - 3)(x + 1)$
Factor the numerator by finding two numbers that multiply to $-3$ and add to $-2$ ; those are $-3$ and $+1$ .
$\frac{(x - 3)(x + 1)}{x + 1} = x - 3 \text{ for } x \neq -1$
Cancel the common factor $(x + 1)$ ; this is valid since we are taking a limit as $x$ approaches $-1$ , not evaluating there directly.
$\lim_{x \to -1} (x - 3) = -1 - 3 = -4$
Substitute $x = -1$ into the simplified expression to find the limit value.

Answer: $\lim_{x \to -1} g(x) = -4$

Factoring the numerator shows that $(x + 1)$ divides both the numerator and denominator, making $x = -1$ a removable discontinuity. The limit is $-4$ . This example demonstrates that removable discontinuities can occur at negative $x$ -values and the limit can be negative, showing that the location and sign of the discontinuity vary with the function.

3. Removable discontinuity in an investment account

Problem

A bank's formula for account balance (in thousands of dollars) at time

n

years is

B(n) = \frac{n^2 - n - 6}{n - 3}

for

n \neq 3

. Year 3 is missing due to a data entry error. What balance does the limit suggest for year 3?

$n^2 - n - 6 = (n - 3)(n + 2)$
Factor the numerator: find two numbers that multiply to $-6$ and add to $-1$ , which are $-3$ and $+2$ .
$\frac{(n - 3)(n + 2)}{n - 3} = n + 2 \text{ for } n \neq 3$
Cancel the common factor $(n - 3)$ ; this works because we are taking a limit as $n$ approaches 3, not evaluating at $n = 3$ directly.
$\lim_{n \to 3} (n + 2) = 3 + 2 = 5$
Substitute $n = 3$ into the simplified expression to find the account value the limit suggests.

Answer: $\lim_{n \to 3} B(n) = 5 \text{ thousand dollars}$

The formula is undefined at $n = 3$ because the record is missing. By factoring and canceling the common factor, we find that the limit as $n$ approaches 3 is 5 thousand dollars. This removable discontinuity is 'removed' by redefining $B(3) = 5$ , restoring the account to its expected balance based on the growth pattern of surrounding years.

Common mistakes

Where Removable Discontinuity usually goes wrong

The mistake	Why it is wrong	The fix
The function is continuous at $x = 2$ because $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$ exists.	Continuity requires that the limit exists AND equals the function value; here $f(2)$ is undefined, so the function is not continuous despite the limit existing.	Verify all three conditions for continuity: (1) $f(a)$ exists, (2) $\lim_{x \to a} f(x)$ exists, and (3) they are equal. If the limit exists but $f(a)$ is undefined or differs, you have a removable discontinuity, not continuity.
Because $f(1)$ is undefined, the limit $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ does not exist.	An undefined function value does not mean the limit is undefined; factoring and canceling can reveal that the limit exists even when the function has a hole.	Factor: $x^2 - 1 = (x - 1)(x + 1)$ . Then $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2$ is well-defined; the discontinuity is removable.
$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ is $\frac{0}{0}$ , so it is undefined.	Getting $\frac{0}{0}$ from direct substitution indicates an indeterminate form requiring further work, not a non-existent limit; algebraic simplification can resolve it.	Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$ . Then $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6$ is well-defined and the discontinuity is removable.

Tips and when to use something else

Always factor or rationalize when direct substitution gives an indeterminate form like $\frac{0}{0}$ ; this is the key method for discovering whether a discontinuity is removable or truly non-existent.
A removable discontinuity is called 'removable' because you can redefine $f(a) = \lim_{x \to a} f(x)$ to make the function continuous; by contrast, jump and infinite discontinuities cannot be removed by redefining a single point.
If the left and right limits both exist but are not equal to each other, you have a jump discontinuity instead of a removable one; always check one-sided limits to distinguish between these cases.
When the numerator and denominator both approach zero, limits by factoring is often quicker than L'Hopital's Rule; try canceling a common factor first before taking derivatives.

Related formulas

Definition of a LimitDefinition of a Limit describes what value a function approaches as its input approaches a specific point—essential for all of calculus.ContinuityContinuity tells you when a function has no jumps or breaks, so you can safely evaluate limits and apply theorems about function behavior.Limits by FactoringLimits by Factoring removes common factors from numerator and denominator to resolve 0/0 indeterminate forms and evaluate limits algebraically.Indeterminate FormsIndeterminate forms are expressions like 0/0 and ∞/∞ that arise when evaluating limits, signaling that direct substitution doesn't work.L'Hopital's RuleL'Hopital's Rule efficiently solves limits that reduce to 0/0 or ∞/∞ forms by taking derivatives of the numerator and denominator separately.

Frequently asked questions

Can I use L'Hopital's Rule to find limits at a removable discontinuity?: Yes, L'Hopital's Rule applies whenever direct substitution yields $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , which is exactly the indeterminate form that appears at removable discontinuities. However, for polynomial and rational functions, factoring and canceling is usually faster and avoids the need to compute derivatives.
Is a removable discontinuity the same as a hole in the graph?: Yes; a removable discontinuity appears as a hole (or open circle) on the graph at the point $(a, L)$ where $L$ is the limit value. The function approaches this point from both sides but never actually reaches it because $f(a)$ is undefined or does not equal $L$ .
How do I know whether a limit is removable or non-existent?: Factor both numerator and denominator; if they share a common factor that cancels, the discontinuity is removable and the limit exists. If no common factor cancels and you still get an indeterminate form, check the left and right limits separately to determine whether the limit truly exists.
What is the difference between a removable discontinuity and a jump discontinuity?: At a removable discontinuity, the limit exists but the function value is missing or incorrect; you can 'fix' this by redefining $f(a)$ . At a jump discontinuity, the left and right limits both exist but differ from each other — no single-point redefinition can make the function continuous because the function jumps.

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Reviewed 2026-09-18