Determining limits using algebraic properties of limits — AP Calculus AB
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: $\lim_{x \to a} [k f(x)] = kL$ 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)
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.
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.
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.
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
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.
Why: Confuses $0/4 = 0$ with $4/0$, which is an undefined infinite limit.
Why: Forgets that you must multiply by $\frac{\text{conjugate}}{\text{conjugate}} = 1$ to keep the expression value the same, leading to an incorrect result.
Why: The difference law only applies when both limits are finite; infinity is not a finite number, so you cannot subtract infinities directly.
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.