Determining limits using algebraic properties of limits

AP Calculus AB· AP Calculus AB CED — Limits and Continuity· 14 min read

1. Basic Algebraic Limit Laws★★☆☆☆⏱ 3 min

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: $lim_{x \to a} [k f (x)] = k L$ for any constant $k$
Sum/difference: $lim_{x \to a} [f (x) \pm g (x)] = L \pm M$
Product: $lim_{x \to a} [f (x) g (x)] = L M$
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 $lim_{x \to a} n f (x) = n L$ (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 $lim_{x \to 2} [5 f (x) + (g (x))^{2} - 2 f (x) + 10]$ .

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.

2. Direct Substitution for Continuous Functions★★☆☆☆⏱ 2 min

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 $lim_{x \to - 1} \frac{x ^{3} - 2 x + 5}{x ^{2} - 3}$ .

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.

3. Factoring to Resolve 0/0 Indeterminate Forms★★★☆☆⏱ 4 min

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 $lim_{x \to 3} \frac{x ^{2} - 5 x + 6}{x ^{2} - 9}$ .

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.

4. Rationalizing to Resolve Indeterminate Forms★★★☆☆⏱ 4 min

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 $lim_{x \to 0} \frac{x + 4 - 2}{x}$ .

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, $x + 4 + 2$ :
$\lim_{x \to 0} \frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)}$
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:
$\frac{1}{\sqrt{4} + 2} = \frac{1}{4}$

✓ Quick check

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

Which of the following is equal to $lim_{x \to - 2} \frac{2 x ^{2} + 3 x - 2}{x ^{2} + 5 x + 6}$ ?
- -5
- -\frac{5}{2}
- \frac{1}{3}
- The limit does not exist

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.

5. Common Pitfalls

Wrong move:

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:

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 move:

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 move:

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:

Confuses $0/4 = 0$ with $4/0$ , which is an undefined infinite limit.

Correct move:

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 move:

When rationalizing, only multiply the numerator by the conjugate, leaving the original denominator unchanged.

Why:

Forgets that you must multiply by $\frac{conjugate}{conjugate} = 1$ to keep the expression value the same, leading to an incorrect result.

Correct move:

Always multiply both numerator and denominator by the conjugate to retain the original value of the expression.

Wrong move:

Applying the difference limit law to $lim_{x \to \infty} (x^{2} - x)$ to get $lim x^{2} - lim x = \infty - \infty = 0$ .

Why:

The difference law only applies when both limits are finite; infinity is not a finite number, so you cannot subtract infinities directly.

Correct move:

Factor or rewrite the expression to resolve the $\infty - \infty$ indeterminate form before applying any limit laws.

Wrong move:

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:

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 move:

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 $a$ .

6. Quick Reference 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	$lim_{x \to a} \frac{f ( x )}{g ( x )} = \frac{l i m f ( x )}{l i m g ( x )}$	Only applies if both limits finite, $lim g (x) \neq = 0$
Power/Root Rule	$lim [f (x)]^{n} = (lim f (x))^{n}$ , $lim n f (x) = n lim f (x)$	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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Evaluate limit of rational function
2022 · FRQ
Limit step for continuity check

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.