Basic Limit Laws

5 min read

Reference: Stewart §2.3

Breaking Limits Into Pieces

What if you could compute limits the same way you compute regular arithmetic? It turns out you can, most of the time.

Consider the limit $\lim_{x \to 3}(x^2 + 5x)$. Rather than doing something complicated, the Limit Laws tell us we can break this into pieces: find the limit of $x^2$, find the limit of $5x$, then add them together. The laws formalize the intuitive idea that “limits respect algebra.”

This is why polynomials are so easy to work with: you can just substitute the value directly, because each piece of the limit works out.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Chapter	1.6
Difficulty	Beginner
Time	~20 minutes

The Limit Laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

Arithmetic Laws

Law	Statement	In Words
Sum	$\lim_{x \to a}[f(x) + g(x)] = L + M$	Limit of sum = sum of limits
Difference	$\lim_{x \to a}[f(x) - g(x)] = L - M$	Limit of difference = difference of limits
Constant Multiple	$\lim_{x \to a}[c \cdot f(x)] = c \cdot L$	Constants factor out
Product	$\lim_{x \to a}[f(x) \cdot g(x)] = L \cdot M$	Limit of product = product of limits
Quotient	$\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{L}{M}$	Limit of quotient = quotient of limits

Important: The Quotient Law requires $M \neq 0$.

Power and Root Laws

Law	Statement	Condition
Power	$\lim_{x \to a}[f(x)]^n = L^n$	$n$ is any positive integer
Root	$\lim_{x \to a}\sqrt[n]{f(x)} = \sqrt[n]{L}$	If $n$ is even, require $L > 0$

Building Block Limits

Limit	Value	Why
$\lim_{x \to a} c$	$c$	Constants don’t change
$\lim_{x \to a} x$	$a$	$x$ approaches $a$
$\lim_{x \to a} x^n$	$a^n$	Power Law applied

Direct Substitution Property

For polynomials and rational functions (when the denominator is nonzero at $a$):

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

Why this works: Polynomials are built from sums, products, and powers, all operations where limits “pass through” unchanged.

When Direct Substitution Works

Can I substitute x = a directly?
         │
         ▼
    Is f(a) defined?
     /          \
   Yes           No
    │             │
    ▼             ▼
  f(a) is      Algebraic work
  the limit    needed first

Practice Problems

Level 1 Direct Polynomial Limit

Evaluate $\lim_{x \to 2}(3x^2 - 4x + 7)$.

Thought Process

This is a polynomial, so direct substitution works. The Sum Law lets me handle each term separately, and the Power Law handles $x^2$.

Just substitute $x = 2$ into the expression.

Show Answer

Since this is a polynomial, we can substitute directly:

$$\lim_{x \to 2}(3x^2 - 4x + 7) = 3(2)^2 - 4(2) + 7 = 12 - 8 + 7 = 11$$

Level 2 Rational Function Limit

Evaluate $\lim_{x \to 3}\frac{x^2 + 2x - 1}{x + 4}$.

Thought Process

First check: is the denominator zero at $x = 3$? We have $3 + 4 = 7 \neq 0$, so the Quotient Law applies.

Use the limit laws: find the limit of the numerator, find the limit of the denominator, then divide.

Show Answer

Check: denominator at $x = 3$ is $3 + 4 = 7 \neq 0$. ✓

By the Quotient Law:

$$\lim_{x \to 3}\frac{x^2 + 2x - 1}{x + 4} = \frac{\lim_{x \to 3}(x^2 + 2x - 1)}{\lim_{x \to 3}(x + 4)} = \frac{9 + 6 - 1}{7} = \frac{14}{7} = 2$$

Level 3 Limit with Radicals

Evaluate $\lim_{x \to 4}\sqrt{2x^2 + x + 3}$.

Thought Process

The Root Law says $\lim \sqrt{f(x)} = \sqrt{\lim f(x)}$, provided the limit inside is positive.

First find the limit of the expression inside the square root, then take the square root.

Show Answer

First, find the limit inside:

$$\lim_{x \to 4}(2x^2 + x + 3) = 2(16) + 4 + 3 = 32 + 4 + 3 = 39$$

Since $39 > 0$, the Root Law applies:

$$\lim_{x \to 4}\sqrt{2x^2 + x + 3} = \sqrt{39}$$

Level 4 Combined Limit Laws

Given that $\lim_{x \to 2} f(x) = 4$ and $\lim_{x \to 2} g(x) = -3$, evaluate:

$$\lim_{x \to 2}\frac{2f(x) - g(x)}{[f(x)]^2 + 1}$$

Thought Process

We’re not given formulas for $f$ and $g$, only their limits. But that’s enough!

Apply the limit laws systematically:

Numerator uses Constant Multiple and Difference Laws
Denominator uses Power and Sum Laws
Then use the Quotient Law

Check that the denominator limit isn’t zero before dividing.

Show Answer

Numerator: $$\lim_{x \to 2}[2f(x) - g(x)] = 2(4) - (-3) = 8 + 3 = 11$$

Denominator: $$\lim_{x \to 2}[f(x)]^2 + 1] = (4)^2 + 1 = 17$$

Since $17 \neq 0$, the Quotient Law applies:

$$\lim_{x \to 2}\frac{2f(x) - g(x)}{[f(x)]^2 + 1} = \frac{11}{17}$$

Level 5 Limit Law Justification

Suppose $\lim_{x \to a} f(x) = L$ exists and $\lim_{x \to a} g(x)$ does not exist.

Can $\lim_{x \to a}[f(x) + g(x)]$ exist? If so, give an example. If not, explain why.
Can $\lim_{x \to a}[f(x) \cdot g(x)]$ exist? If so, give an example. If not, explain why.

Thought Process

This tests whether the limit laws work “backwards”: can the limit of a combination exist even if one piece doesn’t?

For part (a): if $\lim[f + g] = M$ and $\lim f = L$, what would $\lim g$ have to be?

For part (b): think about what happens if $L = 0$. The product $f(x) \cdot g(x)$ might be forced toward 0 even if $g$ oscillates wildly.

Show Answer

(a) No, the sum cannot exist.

Proof by contradiction: Suppose $\lim_{x \to a}[f(x) + g(x)] = M$ exists.

Then by the Difference Law: $$\lim_{x \to a} g(x) = \lim_{x \to a}[(f(x) + g(x)) - f(x)] = M - L$$

This would mean $\lim g(x)$ exists, contradicting our assumption.

(b) Yes, the product can exist when $L = 0$.

Example: Let $f(x) = x$ and $g(x) = \sin(1/x)$ near $x = 0$.

$\lim_{x \to 0} f(x) = 0$ exists
$\lim_{x \to 0} g(x) = \lim_{x \to 0} \sin(1/x)$ does not exist (oscillates)
But $\lim_{x \to 0} x \sin(1/x) = 0$ (by the Squeeze Theorem)

The key insight: When $L = 0$, the product can “absorb” the non-existence of the other limit.

Conceptual Check (CCI-Style)

Conceptual When Laws Fail

True or False: If $\lim_{x \to 3} f(x) = 0$ and $\lim_{x \to 3} g(x) = 0$, then the Quotient Law tells us $\lim_{x \to 3}\frac{f(x)}{g(x)} = 0$.

Thought Process

The Quotient Law says $\lim \frac{f}{g} = \frac{L}{M}$ when $M \neq 0$. What happens when $M = 0$?

Show Answer

False. The Quotient Law requires the denominator limit to be nonzero.

When both limits are 0, we have the indeterminate form $\frac{0}{0}$, and the Quotient Law does not apply. The actual limit could be:

A finite number (e.g., $\lim_{x \to 0}\frac{x}{2x} = \frac{1}{2}$)
Infinity (e.g., $\lim_{x \to 0}\frac{x}{x^2}$ does not exist)
Zero (e.g., $\lim_{x \to 0}\frac{x^2}{x} = 0$)

More algebraic work is needed to determine the limit.

Common Misconceptions

Common misconception

the quotient law applies even when the denominator limit is zero. The quotient law $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f}{\lim g}$ requires $\lim_{x \to a} g(x) \neq 0$. Students sometimes split a fraction and compute numerator and denominator limits separately, then report “undefined” when the denominator limit is zero, without realizing the original limit may still exist. The quotient law fails to apply; a different technique is needed.

Common misconception

the limit of a composition is the composition of the limits. It is tempting to write $\lim_{x \to a} f(g(x)) = f\!\left(\lim_{x \to a} g(x)\right)$ in all cases. This holds when $f$ is continuous at the limit value of $g$, but it fails otherwise. For example, if $g(x) \to 0$ and $f$ has a jump at 0, composing the limits gives the wrong value. The composite limit law requires checking continuity of the outer function at the inner limit.

Mastery Checklist

I can state all seven limit laws (sum, difference, constant multiple, product, quotient, power, root)
I can apply direct substitution to polynomials and rational functions
I recognize when the Quotient Law does NOT apply (denominator limit = 0)
I can compute limits using given information about component function limits
I understand that some limit laws require the individual limits to exist first

Mental Model

Think of limits like a well-behaved calculator:

Just as your calculator can add, subtract, multiply, and divide numbers, limits “respect” these same operations. If $f(x) \to 5$ and $g(x) \to 3$, then $f(x) + g(x) \to 8$ for the same reason that $5 + 3 = 8$.

The only catch: division by zero breaks the calculator; and it breaks the Quotient Law too.

Connections

Looking back:

Limit Intuition gave us the concept; now we have computational tools

Looking ahead:

Indeterminate Forms handles what to do when direct substitution fails
Squeeze Theorem provides a tool for limits the laws can’t handle directly
Continuity formalizes when direct substitution always works

Previous	Up	Next
Limit Intuition	Skills Index	Indeterminate Forms

Last updated: 2026-01-22