Can I use the Sum Law if one of the limits is infinity?

No. The Sum Law requires both limits to be finite real numbers. If \lim g = \infty, the form \lim f + \infty is undefined, and the law does not apply. You may be looking at an indeterminate form like \infty - \infty, which requires a different technique.

What if one limit does not exist at all (not even as infinity)?

The Sum Law cannot be used. If \lim g oscillates or is undefined, you cannot write \lim(f + g) = \lim f + (\text{undefined}). The sum might still have a limit through cancellation, but you must investigate this separately using other methods.

Can I use the Sum Law backward to split a known limit?

Not directly. The law states: if \lim f and \lim g both exist and are finite, then \lim(f + g) = \lim f + \lim g. You cannot reverse this to conclude that knowing \lim(f + g) means you can find \lim f and \lim g separately. Both limits must be verifiable independently first.

Is the Sum Law the same as saying limits are linear?

Yes, in the sense that limits respect sums and scalar multiples—the Sum Law and constant multiple law together make limits a linear operator. This is why calculus can split complex expressions into simpler pieces. However, limits are not linear for products or quotients without additional restrictions, so the Product Law and Quotient Law have different conditions.

Limit Laws

The Sum Law shows that a limit of a sum equals the sum of the limits, provided both limits exist and are finite numbers.

\lim (f \pm g) = \lim f \pm \lim g

Solve a problem with Limit Laws

Type the problem. The solver will use Limit Laws where Limit Laws is the right tool, and tell you when it is not.

What each symbol means

What Limit Laws takes

Symbol	Meaning
$f$	A function representing one piece of the sum; this can be any function, but for the law to apply, $\lim f$ must exist and be a finite real number.
$g$	A second function being added to or subtracted from $f$ ; like $f$ , it must have a finite limit for the law to work.

When to use it

Use the Sum Law when you have a limit of an expression that is clearly a sum or difference of two simpler functions.

Level

Usually taught in: Calculus I

Worked examples

1. Sum of a linear function and a constant

Problem

Evaluate

\lim_{x \to 1} (2x + 4)

$\lim_{x \to 1} (2x + 4) = \lim_{x \to 1} 2x + \lim_{x \to 1} 4$
The Sum Law lets us split the limit of the sum into the sum of the limits, so we write out the limit of each piece separately.
$= 2 \cdot \lim_{x \to 1} x + 4$
The constant multiple law lets us pull a coefficient outside the limit: $\lim_{x \to 1} 2x = 2 \cdot \lim_{x \to 1} x$ . The limit of a constant is the constant itself.
$= 2 \cdot 1 + 4$
Evaluate each limit: $\lim_{x \to 1} x = 1$ and $\lim_{x \to 1} 4 = 4$ .
$= 6$
Simplify the arithmetic.

Answer: $6$

The Sum Law transforms a sum into two simpler limits we can handle separately. This is much faster than trying to compute the limit of $2x + 4$ directly from first principles. Once we split, each piece is straightforward to evaluate.

2. Difference of a polynomial and a rational function

Problem

Evaluate

\lim_{x \to 3} \left( 5x - \frac{1}{x} \right)

$\lim_{x \to 3} \left( 5x - \frac{1}{x} \right) = \lim_{x \to 3} 5x - \lim_{x \to 3} \frac{1}{x}$
The Difference Law (the minus-sign version of the Sum Law) lets us split this into two separate limits.
$= 5 \cdot \lim_{x \to 3} x - \frac{\lim_{x \to 3} 1}{\lim_{x \to 3} x}$
Pull the constant 5 outside the first limit using the constant multiple law. For the second piece, use the quotient law: $\lim \frac{f}{g} = \frac{\lim f}{\lim g}$ (provided $\lim g \ne 0$ ).
$= 5 \cdot 3 - \frac{1}{3}$
Evaluate: $\lim_{x \to 3} x = 3$ , $\lim_{x \to 3} 1 = 1$ , and $\lim_{x \to 3} x = 3$ again.
$= 15 - \frac{1}{3}$
Compute $5 \times 3 = 15$ .
$= \frac{45}{3} - \frac{1}{3} = \frac{44}{3}$
Use a common denominator: convert $15$ to $\frac{45}{3}$ , then subtract to get $\frac{44}{3}$ .

Answer: $\frac{44}{3}$

This example shows the Sum Law working with more complex functions—a polynomial and a rational function. The key insight is that both limits must be finite and real before the law applies. Here, the denominator $\lim_{x \to 3} x = 3 \ne 0$ , so the quotient law (a sibling of the Sum Law) is valid.

3. Basketball scoring: sum of two performance components

Problem

A basketball player's score is modeled as the sum of two components: regular field goals

f(t)

and free-throw bonuses

g(t)

, where

t

is minutes played. As the player approaches 20 minutes of play, their regular scoring approaches

\lim_{t \to 20} f(t) = 12

points, and their free-throw bonus approaches

\lim_{t \to 20} g(t) = 3

points. What total points does the player approach at 20 minutes?

$\lim_{t \to 20} (f(t) + g(t)) = ?$
We need to find the limit of a sum of two functions, which is exactly what the Sum Law handles.
$= \lim_{t \to 20} f(t) + \lim_{t \to 20} g(t)$
Apply the Sum Law to split the limit of the sum into a sum of limits.
$= 12 + 3$
Substitute the given limit values: $\lim_{t \to 20} f(t) = 12$ and $\lim_{t \to 20} g(t) = 3$ .
$= 15$
Add the components to find the total.

Answer: $15 \text{ points}$

This real-world scenario shows why the Sum Law is useful: we model total performance as a sum of independent components, each with its own limit. The Sum Law lets us combine models directly—the total limit is just the sum of the component limits. This is much more efficient than deriving a formula for the combined score and then taking its limit.

Common mistakes

Where Limit Laws usually goes wrong

The mistake	Why it is wrong	The fix
Applying the Sum Law to $\lim_{x \to 0} \left( \sin(x) + \frac{1}{x} \right)$ and writing $0 + \infty$ .	The Sum Law requires both limits to be finite real numbers; it cannot be applied if either limit is $\infty$ , $-\infty$ , or does not exist.	First check that both $\lim f$ and $\lim g$ are finite. Here, $\lim_{x \to 0} \frac{1}{x}$ does not exist, so the Sum Law does not apply; the overall limit is also undefined.
Writing $\lim_{x \to 2} (3x^2 + 4x) = 3(2)^2 + 4(2) = 20$ without showing the Sum Law step.	Skipping the explicit application of the Sum Law makes your work harder to follow and can hide mistakes; the law requires you to split the sum first.	Always show the step: $\lim_{x \to 2} (3x^2 + 4x) = \lim_{x \to 2} 3x^2 + \lim_{x \to 2} 4x = 12 + 8 = 20$ . This makes your reasoning transparent.
Using the Sum Law when told that $\lim f$ or $\lim g$ does not exist, writing $\lim(f + g) = (\text{does not exist}) + 5$ and claiming this equals something.	The Sum Law is a conditional rule: it only applies if the hypotheses (both limits exist and are finite) are satisfied; applying it when they fail is invalid.	If a limit does not exist, the Sum Law does not apply at all. Use other methods like the Squeeze Theorem or L'Hopital's Rule, or re-examine whether the sum might have a limit through cancellation.

Tips and when to use something else

The Sum Law (and the constant multiple law) make limits 'linear' operators—this is why complex limits can be broken into simpler pieces.
For products and quotients, use the Product Law and Quotient Law instead; these have similar forms but require extra care when factors or denominators approach zero.
If one limit fails to exist or is infinite, the Sum Law breaks down completely—try the Squeeze Theorem or L'Hopital's Rule instead.
Always verify that both $\lim f$ and $\lim g$ are finite real numbers before concluding $\lim(f + g) = \lim f + \lim g$ . Forms like $\infty + 5$ or $-\infty + 3$ are undefined and signal the law does not apply.

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.Indeterminate FormsIndeterminate forms are expressions like 0/0 and ∞/∞ that arise when evaluating limits, signaling that direct substitution doesn't work.Limit of a Rational FunctionFind the limit of a rational function at a point by substituting the point directly, provided the denominator is nonzero.ContinuityContinuity tells you when a function has no jumps or breaks, so you can safely evaluate limits and apply theorems about function behavior.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 the Sum Law if one of the limits is infinity?: No. The Sum Law requires both limits to be finite real numbers. If $\lim g = \infty$ , the form $\lim f + \infty$ is undefined, and the law does not apply. You may be looking at an indeterminate form like $\infty - \infty$ , which requires a different technique.
What if one limit does not exist at all (not even as infinity)?: The Sum Law cannot be used. If $\lim g$ oscillates or is undefined, you cannot write $\lim(f + g) = \lim f + (\text{undefined})$ . The sum might still have a limit through cancellation, but you must investigate this separately using other methods.
Can I use the Sum Law backward to split a known limit?: Not directly. The law states: if $\lim f$ and $\lim g$ both exist and are finite, then $\lim(f + g) = \lim f + \lim g$ . You cannot reverse this to conclude that knowing $\lim(f + g)$ means you can find $\lim f$ and $\lim g$ separately. Both limits must be verifiable independently first.
Is the Sum Law the same as saying limits are linear?: Yes, in the sense that limits respect sums and scalar multiples—the Sum Law and constant multiple law together make limits a linear operator. This is why calculus can split complex expressions into simpler pieces. However, limits are not linear for products or quotients without additional restrictions, so the Product Law and Quotient Law have different conditions.

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Reviewed 2026-09-18