Determining limits using algebraic properties of limits — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers the sum, difference, product, quotient, constant multiple, power, and root limit properties, direct substitution for continuous functions, factoring, and rationalizing to resolve 0/0 indeterminate limits.

You should already know: Basic definition of one-sided and two-sided limits, polynomial and rational function algebra, definition of continuity at a point.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Determining limits using algebraic properties of limits?

Determining limits using algebraic properties of limits is the core set of algebraic rules that let you compute the limit of a combination of functions from the known limits of the individual component functions. Unlike numerical estimation or graphical approximation, algebraic methods give exact, provable limit values, which is required for full credit on most AP Calculus AB problems that involve limits. Per the AP Calculus AB Course and Exam Description (CED), this topic is a core part of Unit 1, which accounts for 10-12% of the total exam score, with this specific topic making up roughly 3-4% of the exam. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a foundational step for problems involving continuity, the definition of the derivative, or asymptotic behavior of functions. Synonyms for this topic include limit laws, algebraic limit evaluation, and limit arithmetic.

2. Basic Algebraic Limit Laws

The basic algebraic limit laws (also called limit properties) are the fundamental rules that govern how we combine limits of known functions to get the limit of a more complex function. These rules only apply when both ${\lim }_{x\to a}f(x)=L$ and ${\lim }_{x\to a}g(x)=M$ exist as finite real numbers. The key rules are:

• Constant multiple: ${\lim }_{x\to a}[kf(x)]=kL$ 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 $M\ne 0$
• 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 $L\ge 0$ if $n$ is even)

Intuition for these rules is straightforward: if $f(x)$ gets arbitrarily close to $L$ and $g(x)$ gets arbitrarily close to $M$ near $x=a$, then their combinations get arbitrarily close to the corresponding combination of $L$ and $M$.

Worked Example

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

1. Apply the sum/difference law to split the expression into three separate limits, since both original limits are finite: $$\underset{x\to 2}{\lim }5f(x)+\underset{x\to 2}{\lim }(g(x){)}^2-\underset{x\to 2}{\lim }\sqrt{2f(x)+10}$$
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 other rules: $\sqrt{{\lim }_{x\to 2}(2f(x)+10)}=\sqrt{2(-3)+10}=\sqrt{4}=2$
5. Combine the results: $-15+16-2=-1$

The final limit is $\boxed{-1}$.

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.

3. Direct Substitution for Continuous Functions

All basic continuous functions (polynomials, rational functions, roots, exponentials, and trigonometric functions) are continuous at every point in their domains. By the definition of continuity at a point $a$, ${\lim }_{x\to a}f(x)=f(a)$, which means we can just substitute $x=a$ directly into the function to get the limit. This result comes directly from the basic limit laws: if each component function of $f(x)$ is continuous, the combination rules give that the whole function is continuous, so direct substitution works.

Direct substitution is the fastest method for evaluating limits when it works. It only fails when $f(a)$ is undefined (e.g., $0/0$, non-zero over zero, which correspond to points outside the domain of the function). When that happens, we need to use additional algebraic methods to resolve the indeterminate or undefined form.

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 $x=-1$: the only discontinuities of this rational function occur when the denominator is zero.
2. Calculate the denominator at $x=-1$: $(-1{)}^2-3=1-3=-2\ne 0$, so $x=-1$ is in the domain of the function, so direct substitution is valid.
3. Substitute $x=-1$ into the numerator: $(-1{)}^3-2(-1)+5=-1+2+5=6$.
4. Apply the quotient rule to get the final limit: $\frac{6}{-2}=-3$.

The final limit is $\boxed{-3}$.

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 (it approaches $\pm \infty$), so do not try to "cancel" anything.

4. Factoring to Resolve 0/0 Indeterminate Forms

The most common indeterminate form you will encounter on the AP exam is $0/0$, which occurs when direct substitution gives zero in both the numerator and denominator of a rational function. This $0/0$ form tells you that $(x-a)$ is a common factor of both the numerator and denominator, where $a$ is the point you are approaching. We can cancel this common factor because when we take the limit as $x\to a$, we only care about values of $x$ near $a$, not at $a$, so $(x-a)\ne 0$, and canceling is algebraically valid. After canceling, the simplified function is continuous at $a$, so we can use direct substitution to find the limit.

Worked Example

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

1. Test direct substitution: numerator at $x=3$ is $3^2-5(3)+6=9-15+6=0$, denominator is $3^2-9=0$, so we have a $0/0$ indeterminate form that requires factoring.
2. Factor both the numerator and denominator: $x^2-5x+6=(x-2)(x-3)$ and $x^2-9=(x-3)(x+3)$.
3. Cancel the common $(x-3)$ factor, which is valid since $x\ne 3$ for the limit: $$\underset{x\to 3}{\lim }\frac{(x-2)(x-3)}{(x-3)(x+3)}=\underset{x\to 3}{\lim }\frac{x-2}{x+3}$$
4. Use direct substitution on the simplified function: $\frac{3-2}{3+3}=\frac{1}{6}$.

The final limit is $\boxed{\frac{1}{6}}$.

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.

5. Rationalizing to Resolve Indeterminate Forms

Another common case of $0/0$ indeterminate forms involves square roots in the numerator or denominator, where factoring is not immediately possible. In this case, we use the method of rationalizing, which relies on the difference of squares identity: $(a-b)(a+b)=a^2-b^2$. We multiply both the numerator and denominator by the conjugate of the radical expression (the same expression with the sign flipped on the radical term). This eliminates the radical from the numerator or denominator, reveals a common factor that we can cancel, and lets us use direct substitution to find the limit.

Worked Example

Evaluate ${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}$.

1. Test direct substitution: numerator at $x=0$ is $\sqrt{0+4}-2=0$, denominator is $0$, so we have a $0/0$ indeterminate form that requires rationalizing.
2. Multiply both the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+4}+2$: $$\underset{x\to 0}{\lim }\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}$$
3. Simplify the numerator using the difference of squares: $(\sqrt{x+4}{)}^2-(2{)}^2=(x+4)-4=x$. This gives: $$\underset{x\to 0}{\lim }\frac{x}{x(\sqrt{x+4}+2)}$$
4. Cancel the common $x$ factor, which is valid since $x\ne 0$ when taking the limit. This leaves: $$\underset{x\to 0}{\lim }\frac{1}{\sqrt{x+4}+2}$$
5. Use direct substitution to get the final result: $\frac{1}{\sqrt{4}+2}=\frac{1}{4}$.

The final limit is $\boxed{\frac{1}{4}}$.

Exam tip: Always multiply both numerator and denominator by the conjugate to keep the value of the expression the same; many students forget to multiply the denominator, leading to an incorrect answer.

6. Common Pitfalls (and how to avoid them)

• Wrong move: After getting ${\lim }_{x\to 3}\frac{(x-2)(x-3)}{(x-3)(x+3)}$, you plug in $x=3$ before canceling $(x-3)$, get $0/0$ and conclude the limit does not exist. Why: Students confuse the value of the function at $x=a$ with the limit as $x$ approaches $a$; they think $0/0$ is undefined so 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: Students confuse $0/4$ (which is 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 (or does not exist as a finite real number), do not invert the fraction.
• Wrong move: When rationalizing ${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}$, only multiply the numerator by the conjugate, leaving the original denominator unchanged, leading to a final answer of $\frac{1}{4x}$ which is incorrectly identified as infinite. Why: Students forget that you must multiply by $\frac{\text{conjugate}}{\text{conjugate}}=1$ to keep the expression value the same. 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 $x=2$, you plug in $f(2)g(2)$ to get the wrong limit. Why: Students assume 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 $a$.

7. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is equal to ${\lim }_{x\to -2}\frac{2x^2+3x-2}{x^2+5x+6}$? A) $-5$ B) $-\frac{5}{2}$ C) $\frac{1}{3}$ D) The limit does not exist

Worked Solution: First, test direct substitution at $x=-2$: the numerator evaluates to $2(-2{)}^2+3(-2)-2=0$ and the denominator evaluates to $(-2{)}^2+5(-2)+6=0$, giving the indeterminate $0/0$ form. Factor both polynomials: the numerator factors to $(2x-1)(x+2)$ and the denominator factors to $(x+2)(x+3)$. The common $(x+2)$ factor can be canceled since $x\ne -2$ when taking the limit, leaving ${\lim }_{x\to -2}\frac{2x-1}{x+3}$. Direct substitution gives $\frac{2(-2)-1}{-2+3}=\frac{-5}{1}=-5$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\frac{\sqrt{x+7}-3}{x^2-2x-3}$. (a) Find ${\lim }_{x\to 2}f(x)$ using algebraic limit properties. Show all steps. (b) Find ${\lim }_{x\to 2}[3(f(x){)}^2-5f(x)]$ using your answer from part (a) and the limit properties. (c) Is the function $g(x)=\{\begin{array}{ll}f(x)& x\ne 2\\ \text{your answer from (a)}& x=2\end{array}$ continuous at $x=2$? Justify your answer using the definition of continuity.

Worked Solution: (a) Test direct substitution at $x=2$: numerator is $\sqrt{2+7}-3=3-3=0$, denominator is $2^2-2(2)-3=-3\ne 0$. The function is continuous at $x=2$, so direct substitution gives ${\lim }_{x\to 2}f(x)=\frac{0}{-3}=0$. $\boxed{\underset{x\to 2}{\lim }f(x)=0}$

(b) Apply the constant multiple, power, and difference limit laws: $$\underset{x\to 2}{\lim }[3(f(x){)}^2-5f(x)]=3{\left(\underset{x\to 2}{\lim }f(x)\right)}^2-5\left(\underset{x\to 2}{\lim }f(x)\right)$$ Substitute ${\lim }_{x\to 2}f(x)=0$ to get $3(0{)}^2-5(0)=0$. $\boxed{0}$

(c) A function is continuous at $x=a$ if three conditions are met: (1) $g(a)$ is defined, (2) ${\lim }_{x\to a}g(x)$ exists, and (3) ${\lim }_{x\to a}g(x)=g(a)$. For $g(x)$ at $x=2$: $g(2)=0$ is defined, ${\lim }_{x\to 2}g(x)={\lim }_{x\to 2}f(x)=0$ exists, and ${\lim }_{x\to 2}g(x)=g(2)$. All three conditions are satisfied, so $g(x)$ is continuous at $x=2$.

Question 3 (Application / Real-World Style)

The position of an object moving along a line is given by $s(t)=\frac{t^2}{\sqrt{t+9}-3}$ meters, where $t$ is time in seconds for $t>0$. The average velocity of the object over the interval $[0,t]$ is $v_{\text{avg}}(t)=\frac{s(t)-s(0)}{t-0}$. The instantaneous velocity at $t=0$ is defined as the limit of the average velocity as $t\to 0^{+}$. Find the instantaneous velocity at $t=0$ using algebraic limit properties.

Worked Solution: First, confirm $s(0)=0$ (the limit of $s(t)$ as $t\to 0$ is 0, so $s(0)=0$ for this position function). Then simplify the average velocity: $$v_{\text{avg}}(t)=\frac{\frac{t^2}{\sqrt{t+9}-3}-0}{t}=\frac{t}{\sqrt{t+9}-3}$$ We need ${\lim }_{t\to 0^{+}}\frac{t}{\sqrt{t+9}-3}$, which is a $0/0$ indeterminate form. Multiply numerator and denominator by the conjugate $\sqrt{t+9}+3$: $$\underset{t\to 0^{+}}{\lim }\frac{t(\sqrt{t+9}+3)}{(t+9)-9}=\underset{t\to 0^{+}}{\lim }\frac{t(\sqrt{t+9}+3)}{t}=\underset{t\to 0^{+}}{\lim }(\sqrt{t+9}+3)$$ Direct substitution gives $\sqrt{9}+3=6$. The instantaneous velocity at $t=0$ is 6 meters per second. In context, this means the object is moving to the right at 6 m/s at the moment $t=0$.

8. Quick Reference Cheatsheet

9. What's Next

This chapter is the foundational prerequisite for all upcoming limit topics and most of calculus. Immediately after mastering algebraic limit properties, you will learn to evaluate limits at infinity to find horizontal asymptotes, and to analyze continuity of functions at a point. Without being able to accurately compute limits using algebraic properties, you will not be able to correctly identify discontinuities, confirm continuity for the Intermediate Value Theorem, or compute derivatives using the limit definition in the next unit. This topic feeds into all core concepts of AP Calculus AB: every derivative and definite integral is defined as a limit, so algebraic limit evaluation is a required step for almost all non-trivial problems on the exam.

Evaluating limits at infinity for asymptotes Continuity at a point and types of discontinuity Definition of the derivative as a limit Intermediate Value Theorem (IVT) application