# Determining limits using algebraic properties of limits

> AP Calculus AB · Unit 1: Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u1-determining-limits-using-algebraic-properties/

This module covers core limit properties: sum, difference, product, quotient, constant multiple, power, and root rules, plus direct substitution, factoring, and rationalizing to resolve 0/0 indeterminate limits, a core AP Calculus AB skill.

**Prerequisites:** Basic definition of one-sided and two-sided limits; Polynomial and rational function algebra; [Definition of continuity at a point](https://www.owlsprep.com/study/ap-calculus-ab-u1-definition-of-continuity/)

## Learning objectives

- Apply basic algebraic limit laws to evaluate limits of function combinations
- Use direct substitution to evaluate limits of continuous functions
- Resolve 0/0 indeterminate forms using factoring
- Resolve 0/0 indeterminate forms involving radicals via rationalizing

## Basic Algebraic Limit Laws

The basic algebraic limit laws (or limit properties) are fundamental rules that let you combine known limits of individual functions to evaluate the limit of a more complex expression. All laws below require that both $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ exist as finite real numbers.

- Constant multiple: $\lim_{x \to a} [k f(x)] = kL$ for any constant $k$
- Sum/difference: $\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$
- Product: $\lim_{x \to a} [f(x)g(x)] = LM$
- Quotient: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ for $M \neq 0$
- Power/root: $\lim_{x \to a} [f(x)]^n = L^n$ for any positive integer $n$, and $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (with $L \geq 0$ if $n$ is even)

**Worked example:** Given that $\lim_{x \to 2} f(x) = -3$ and $\lim_{x \to 2} g(x) = 4$, find $\lim_{x \to 2} \left[ 5f(x) + (g(x))^2 - \sqrt{2 f(x) + 10} \right]$.

1. Apply the sum/difference law to split the expression into three separate limits, since both original limits are finite:

   $$lim_{x \to 2} 5f(x) + \lim_{x \to 2} (g(x))^2 - \lim_{x \to 2} \sqrt{2 f(x) + 10}$$
2. Apply the constant multiple rule to the first term:

   $$5 \cdot \lim_{x \to 2} f(x) = 5(-3) = -15$$
3. Apply the power rule to the second term:

   $$\left(\lim_{x \to 2} g(x)\right)^2 = 4^2 = 16$$
4. Simplify the radical term using the sum, constant multiple and root rules:

   $$\sqrt{\lim_{x \to 2} (2f(x) + 10)} = \sqrt{2(-3) + 10} = \sqrt{4} = 2$$
5. Combine the results to get the final limit:

   $$-15 + 16 - 2 = -1$$

> **Exam tip:** Always confirm that both original limits exist as finite values before applying the limit laws; you cannot apply the quotient law or difference law to infinite limits without adjusting for indeterminate forms first.

## Direct Substitution for Continuous Functions

All basic continuous functions (polynomials, rational functions, roots, exponentials, trigonometric functions) are continuous at every point in their domains. By definition of continuity at $a$, $\lim_{x \to a} f(x) = f(a)$, so you can substitute $x=a$ directly into the function to get the exact limit. This method is the fastest way to evaluate limits when valid, but fails if $f(a)$ is undefined.

**Worked example:** Evaluate $\lim_{x \to -1} \frac{x^3 - 2x + 5}{x^2 - 3}$.

1. Check if the function is defined and continuous at $x=-1$: this rational function only has discontinuities where the denominator is zero. Calculate denominator at $x=-1$:

   $$(-1)^2 - 3 = 1 - 3 = -2 \neq 0$$
2. Since $x=-1$ is in the domain, direct substitution is valid. Substitute $x=-1$ into the numerator:

   $$(-1)^3 - 2(-1) + 5 = -1 + 2 + 5 = 6$$
3. Apply the quotient rule to get the final limit:

   $$\frac{6}{-2} = -3$$

> **Exam tip:** Always check the denominator before using direct substitution for rational functions; if the denominator is zero and the numerator is non-zero, the limit does not exist as a finite value, so do not try to cancel anything.

## Factoring to Resolve 0/0 Indeterminate Forms

The most common indeterminate form on the AP exam is $0/0$, which occurs when direct substitution gives zero in both numerator and denominator of a rational function. This means $(x-a)$ is a common factor of both numerator and denominator, where $a$ is the point you are approaching. Because when taking $\lim_{x \to a}$, we only care about values near $a$ (not at $a$), $(x-a) \neq 0$, so canceling the common factor is algebraically valid. After canceling, use direct substitution on the simplified expression.

**Worked example:** Evaluate $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 9}$.

1. Test direct substitution at $x=3$:

   $$3^2 - 5(3) + 6 = 0, \quad 3^2 - 9 = 0$$
2. We have a $0/0$ indeterminate form, so we factor both numerator and denominator:

   $$x^2 - 5x + 6 = (x-2)(x-3), \quad x^2 - 9 = (x-3)(x+3)$$
3. Cancel the common $(x-3)$ factor (valid since $x \neq 3$ when taking the limit):

   $$\lim_{x \to 3} \frac{(x-2)(x-3)}{(x-3)(x+3)} = \lim_{x \to 3} \frac{x-2}{x+3}$$
4. Use direct substitution on the simplified function:

   $$\frac{3-2}{3+3} = \frac{1}{6}$$

> **Exam tip:** When you get 0/0 after direct substitution, do not conclude the limit is 0 or does not exist; always check for common factors first. Most AP MCQ 0/0 problems have a finite limit from factoring.

## Rationalizing to Resolve Indeterminate Forms

When you get a $0/0$ indeterminate form with square roots, factoring is not immediately possible. Instead, use rationalizing, which relies on the difference of squares identity: $(a-b)(a+b) = a^2 - b^2$. Multiply both numerator and denominator by the conjugate of the radical expression (flip the sign on the radical term) to eliminate the radical, reveal a common factor to cancel, then use direct substitution.

**Worked example:** Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$.

1. Test direct substitution at $x=0$:

   $$\sqrt{0+4} - 2 = 0, \quad \text{denominator } = 0$$
2. We have a $0/0$ indeterminate form that requires rationalizing. Multiply numerator and denominator by the conjugate of the numerator, $\sqrt{x+4} + 2$:

   $$\lim_{x \to 0} \frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)}$$
3. Simplify the numerator using difference of squares:

   $$(\sqrt{x+4})^2 - (2)^2 = (x+4) - 4 = x$$
4. This simplifies the expression, then cancel the common $x$ factor (valid since $x \neq 0$ when taking the limit):

   $$\lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}$$
5. Use direct substitution to get the final result:

   $$\frac{1}{\sqrt{4} + 2} = \frac{1}{4}$$

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. Which of the following is equal to $\lim_{x \to -2} \frac{2x^2 + 3x - 2}{x^2 + 5x + 6}$?

   - -5
   - -\frac{5}{2}
   - \frac{1}{3}
   - The limit does not exist

   *Answer:* -5

   *Why:* Correct! Direct substitution gives 0/0, factoring cancels the common $(x+2)$ term, and direct substitution on the simplified expression gives a final limit of $-5$.

> **Exam tip:** Always multiply both numerator and denominator by the conjugate to keep the value of the expression the same; forgetting to multiply the denominator is a common mistake that leads to wrong answers.

## Common pitfalls

- **Wrong:** After getting $\lim_{x \to 3} \frac{(x-2)(x-3)}{(x-3)(x+3)}$, substitute $x=3$ before canceling, get $0/0$ and conclude the limit does not exist.
  - Why it fails: Confuses the value of the function at $x=a$ with the limit as $x$ approaches $a$; assumes $0/0$ undefined means the limit is undefined.
  - Correct: Always cancel common factors first, since for the limit $x$ never actually equals $a$, so $(x-a)$ is non-zero and can be canceled.
- **Wrong:** Applying the quotient limit law to $\lim_{x \to 2} \frac{x+2}{x-2}$ by plugging in 2, getting $4/0$ and writing the limit as $0$.
  - Why it fails: Confuses $0/4 = 0$ with $4/0$, which is an undefined infinite limit.
  - Correct: When direct substitution gives non-zero numerator over zero denominator, recognize the limit is infinite (does not exist as a finite real number), do not invert the fraction.
- **Wrong:** When rationalizing, only multiply the numerator by the conjugate, leaving the original denominator unchanged.
  - Why it fails: Forgets that you must multiply by $\frac{\text{conjugate}}{\text{conjugate}} = 1$ to keep the expression value the same, leading to an incorrect result.
  - Correct: Always multiply both numerator and denominator by the conjugate to retain the original value of the expression.
- **Wrong:** Applying the difference limit law to $\lim_{x \to \infty} (x^2 - x)$ to get $\lim x^2 - \lim x = \infty - \infty = 0$.
  - Why it fails: The difference law only applies when both limits are finite; infinity is not a finite number, so you cannot subtract infinities directly.
  - Correct: Factor or rewrite the expression to resolve the $\infty - \infty$ indeterminate form before applying any limit laws.
- **Wrong:** When evaluating $\lim_{x \to 2} f(x)g(x)$ where $f$ is discontinuous at $x=2$, plug in $f(2)g(2)$ to get the limit.
  - Why it fails: Assumes all functions are continuous everywhere, but $f(a)$ may not equal the limit of $f(x)$ as $x$ approaches $a$ for discontinuous functions.
  - Correct: If you are not given that $f$ and $g$ are continuous at $a$, use the given limits of $f$ and $g$ when applying product law, not the function values at $a$.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Constant Multiple Rule | $\lim_{x \to a} k f(x) = k \lim_{x \to a} f(x)$ | Applies if $\lim_{x \to a} f(x)$ exists |
| Sum/Difference Rule | $\lim_{x \to a} [f(x) \pm g(x)] = \lim f(x) \pm \lim g(x)$ | Only applies if both limits are finite |
| Product Rule | $\lim_{x \to a} [f(x)g(x)] = (\lim f(x))(\lim g(x))$ | Only applies if both limits are finite |
| Quotient Rule | $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$ | Only applies if both limits finite, $\lim g(x) \neq 0$ |
| Power/Root Rule | $\lim [f(x)]^n = (\lim f(x))^n$, $\lim \sqrt[n]{f(x)} = \sqrt[n]{\lim f(x)}$ | Root rule requires $\lim f(x) \geq 0$ if $n$ even |
| Direct Substitution | $\lim_{x \to a} f(x) = f(a)$ | Only applies if $f$ continuous at $x=a$, $f(a)$ defined |
| Factoring for 0/0 | Cancel common $(x-a)$ factor, evaluate simplified limit | Valid because $x \neq a$ when taking $\lim_{x \to a}$ |
| Rationalizing for 0/0 | Multiply numerator/denominator by conjugate of radical | Eliminates radical to reveal common factor for canceling |

## What's next

Mastering algebraic limit properties is a foundational prerequisite for all upcoming topics in AP Calculus AB. This skill is required to analyze continuity, apply the Intermediate Value Theorem, compute derivatives using the limit definition, and evaluate integrals, since all derivatives and integrals are defined as limits. After mastering this sub-topic, you will next move to evaluating limits at infinity to find horizontal asymptotes, followed by formal analysis of continuity at a point and over intervals. Without solid proficiency in algebraic limit evaluation, you will struggle with almost all non-trivial problems on the AP exam, as it is a required intermediate step for most FRQ and many MCQ questions.

- [Determining limits using algebraic manipulation](https://www.owlsprep.com/study/ap-calculus-ab-u1-determining-limits-using-algebraic-manipulation/)
- [Selecting procedures for determining limits](https://www.owlsprep.com/study/ap-calculus-ab-u1-selecting-procedures-for-determining-limits/)
- [Squeeze Theorem](https://www.owlsprep.com/study/ap-calculus-ab-u1-squeeze-theorem/)

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