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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Basic limit laws (sum, difference, product, quotient, power, root), direct substitution for continuous functions, factoring/canceling for 0/0 indeterminate forms, rationalizing, and algebraic evaluation of one-sided limits for piecewise functions.

You should already know: Basic limit notation and the informal definition of a limit. Algebraic manipulation of polynomials, rational functions, and radical expressions. One-sided limit notation and existence conditions for two-sided limits.

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 BC 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 (often called the limit laws) is a core topic in Unit 1: Limits and Continuity, which makes up ~10-12% of the AP Calculus BC exam per the College Board CED. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone skill or as a step in larger problems involving continuity, derivatives, or integrals. Unlike graphical or numerical estimation, algebraic properties of limits allow us to compute exact limit values, which is required for full credit on most exam questions. The core notation used is ${\lim }_{x\to a}f(x)$, representing the value $f(x)$ approaches as $x$ gets arbitrarily close to (but does not equal) $a$, and all properties here apply equally to two-sided limits as $x\to a$ and one-sided limits as $x\to a^{-}$ or $x\to a^{+}$. Mastery of this topic is non-negotiable: nearly every subsequent limit-based skill in the course builds on these algebraic rules.

2. Basic Limit Laws and Direct Substitution

The fundamental algebraic properties of limits, called limit laws, are rules that let us break complex limits into simpler, solvable parts. The rules rely on a core assumption: that ${\lim }_{x\to a}f(x)=L$ and ${\lim }_{x\to a}g(x)=M$ both exist (are finite real numbers). The key laws are:

1. Constant multiple: ${\lim }_{x\to a}cf(x)=cL$ for any constant $c$
2. Sum/difference: ${\lim }_{x\to a}[f(x)\pm g(x)]=L\pm M$
3. Product: ${\lim }_{x\to a}[f(x)g(x)]=LM$
4. Quotient: ${\lim }_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}$, if and only if $M\ne 0$
5. Power/root: ${\lim }_{x\to a}[f(x){]}^n=L^n$ and ${\lim }_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}$ (for even $n$, $L\ge 0$)

For any function that is continuous at $x=a$ (which is true for all polynomials, rational functions, trigonometric, exponential, and logarithmic functions on their domains), we can use direct substitution: ${\lim }_{x\to a}f(x)=f(a)$. This works because continuity is defined as the limit equaling the function value.

Worked Example

Problem: Evaluate ${\lim }_{x\to 2}\left(3x^2-5x+\sqrt{2x+1}\right)$ using algebraic limit properties.

1. First, confirm the function is defined at $x=2$: the expression inside the square root is $2(2)+1=5>0$, so $x=2$ is in the domain of the function, and direct substitution applies.
2. Split the limit using sum/difference and constant multiple rules: $$\underset{x\to 2}{\lim }\left(3x^2-5x+\sqrt{2x+1}\right)=3{\left(\underset{x\to 2}{\lim }x\right)}^2-5\left(\underset{x\to 2}{\lim }x\right)+\sqrt{\underset{x\to 2}{\lim }(2x+1)}$$
3. Use the basic identity ${\lim }_{x\to 2}x=2$ and substitute: $$=3(2{)}^2-5(2)+\sqrt{2(2)+1}=12-10+\sqrt{5}=2+\sqrt{5}$$
4. The final limit is $2+\sqrt{5}$.

Exam tip: Always confirm the evaluation point is in the function's domain before using direct substitution—if it is, no extra work is needed to find the limit.

3. Factoring and Canceling for 0/0 Indeterminate Forms

When you substitute $x=a$ into a rational function and get $\frac{0}{0}$, you have an indeterminate form: this does not mean the limit does not exist, it just means the quotient law cannot be applied directly. $\frac{0}{0}$ almost always means the numerator and denominator share a common factor of $(x-a)$, which causes both to equal zero at $x=a$. Because we are taking the limit as $x\to a$, $x$ never actually equals $a$, so we can safely cancel the common $(x-a)$ factor, then use direct substitution on the simplified expression. This method is the most common way to solve 0/0 indeterminate rational function limits on the AP exam.

Worked Example

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

1. First test direct substitution: numerator at $x=-3$ is $(-3{)}^2+(-3)-6=9-3-6=0$, denominator is $(-3{)}^2-9=9-9=0$, so we have a 0/0 indeterminate form, so we use factoring.
2. Factor both the numerator and denominator: $x^2+x-6=(x+3)(x-2)$ and $x^2-9=(x+3)(x-3)$
3. Cancel the common factor $(x+3)$, which is valid because $x\to -3$, so $x\ne -3$ and $(x+3)\ne 0$. This simplifies the limit to: $$\underset{x\to -3}{\lim }\frac{x-2}{x-3}$$
4. Use direct substitution on the simplified expression: $\frac{-3-2}{-3-3}=\frac{-5}{-6}=\frac{5}{6}$. The limit equals $\frac{5}{6}$.

Exam tip: If you get 0/0 after substitution, always look for a common linear factor first—9 times out of 10 on the AP exam, that factor will cancel out cleanly with no messy factoring required.

4. Rationalizing for Indeterminate Forms

When 0/0 indeterminate forms involve radicals, factoring will not work directly, so we use the method of rationalizing. This 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 conjugate changes the sign between the radical term and the constant term, not the sign inside the radical). This eliminates the radical in the numerator or denominator, reveals a common factor that causes the 0/0, and lets us cancel and substitute, just like with factoring. This method is most commonly used for limits with square roots in the numerator.

Worked Example

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

1. Test direct substitution: $\frac{\sqrt{0+4}-2}{0}=\frac{2-2}{0}=\frac{0}{0}$, so indeterminate. Factoring is not possible here due to the radical, so we use rationalizing.
2. Multiply numerator and denominator by the conjugate of the numerator, $\sqrt{x+4}+2$: $$\underset{x\to 0}{\lim }\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}$$
3. Multiply out the numerator using difference of squares: $(x+4)-(2{)}^2=x+4-4=x$. This simplifies the expression to: $$\underset{x\to 0}{\lim }\frac{x}{x(\sqrt{x+4}+2)}$$
4. Cancel the common $x$ factor (valid because $x\to 0$, so $x\ne 0$), leaving ${\lim }_{x\to 0}\frac{1}{\sqrt{x+4}+2}$.
5. Direct substitute: $\frac{1}{\sqrt{4}+2}=\frac{1}{4}$. The limit equals $\frac{1}{4}$.

Exam tip: Always place the conjugate on the side that contains the radical—if the radical is in the numerator, multiply by the numerator's conjugate; if it is in the denominator, use the denominator's conjugate.

5. Algebraic Evaluation of Limits for Piecewise Functions

To find the limit as $x$ approaches a point where a piecewise function changes its rule, you evaluate the left-hand limit ($x\to a^{-}$) using the rule that applies for $x<a$, and the right-hand limit ($x\to a^{+}$) using the rule that applies for $x>a$. You then use the two-sided limit existence rule: ${\lim }_{x\to a}f(x)$ exists if and only if ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)$. You use the same algebraic properties (direct substitution, factoring, rationalizing) to evaluate each one-sided limit separately. This is a very common MCQ question on the AP exam, often paired with a question about continuity.

Worked Example

Problem: Let $f(x)=\{\begin{array}{ll}\frac{x^2-4}{x-2}& x<2\\ ax+6& x\ge 2\end{array}$. Find the value of $a$ such that ${\lim }_{x\to 2}f(x)$ exists.

1. Evaluate the left-hand limit, $x\to 2^{-}$, so use the first rule for $x<2$: ${\lim }_{x\to 2^{-}}\frac{x^2-4}{x-2}$. Direct substitution gives 0/0, so factor: $\frac{(x-2)(x+2)}{x-2}=x+2$ after canceling $(x-2)$. The left-hand limit is $2+2=4$.
2. Evaluate the right-hand limit, $x\to 2^{+}$, so use the second rule for $x\ge 2$: ${\lim }_{x\to 2^{+}}ax+6=2a+6$ by direct substitution.
3. Set left-hand limit equal to right-hand limit for the two-sided limit to exist: $4=2a+6$. Solve for $a$: $2a=-2$, so $a=-1$.

Exam tip: Always double-check which piece corresponds to which side: $a^{-}$ means $x<a$, so use the rule for $x<a$, not the reverse.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Concluding a limit does not exist immediately after getting 0/0 from direct substitution. Why: Students confuse the function being undefined at $a$ with the limit not existing at $a$. 0/0 is an indeterminate form, not a conclusion. Correct move: If you get 0/0, always proceed to factoring or rationalizing to simplify the expression before concluding the limit does not exist.
• Wrong move: Canceling the $(x-a)$ factor and then concluding $f(a)$ equals the simplified value. Why: Students confuse the limit as $x\to a$ with the value of the function at $a$. Correct move: Explicitly note that $x\ne a$ when canceling, so the equality only holds for the limit, not the function value at $a$.
• Wrong move: Using the quotient law when the denominator limit is zero, without checking the numerator limit. Why: Students memorize the quotient law's denominator restriction and forget that 0/0 is indeterminate, while non-zero/zero has no finite limit. Correct move: If the denominator limit is zero, check the numerator limit first: if it is also zero, simplify; if not, the limit does not exist (or is infinite).
• Wrong move: Using the wrong conjugate by changing the sign inside the radical instead of between the terms. For example, writing the conjugate of $\sqrt{x+4}-2$ as $\sqrt{x-4}+2$. Why: Students confuse the position of the sign in the expression. Correct move: Remember that the conjugate of $A-B$ is $A+B$, so only flip the sign between the two terms, leave the radical term unchanged.
• Wrong move: Evaluating the wrong one-sided limit for piecewise functions, using the $x\ge a$ rule for $x\to a^{-}$. Why: Students mix up the notation for left and right limits. Correct move: Write a quick reminder next to the limit: $a^{-}$ = less than $a$, $a^{+}$ = greater than $a$, then match to the correct piece rule.

7. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is equal to ${\lim }_{x\to 4}\frac{2x-8}{x^2-3x-4}$? A) $\frac{2}{5}$ B) $\frac{1}{2}$ C) $0$ D) The limit does not exist

Worked Solution: First test direct substitution: at $x=4$, the numerator is $2(4)-8=0$, and the denominator is $4^2-3(4)-4=16-12-4=0$, so we have a 0/0 indeterminate form. Factor the numerator and denominator: the numerator becomes $2(x-4)$, and the denominator factors to $(x-4)(x+1)$. Cancel the common $(x-4)$ factor (valid because $x\ne 4$ as $x\to 4$), leaving ${\lim }_{x\to 4}\frac{2}{x+1}$. Substitute $x=4$ to get $\frac{2}{4+1}=\frac{2}{5}$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\{\begin{array}{ll}\frac{\sqrt{9+x}-3}{2x}& x<0\\ \frac{3x+a}{x+2}& x\ge 0\end{array}$ (a) Evaluate ${\lim }_{x\to 0^{-}}f(x)$ using algebraic properties of limits. (b) Find the value of $a$ such that ${\lim }_{x\to 0}f(x)$ exists. (c) Given that ${\lim }_{x\to 0}f(x)$ exists, what value of $a$ makes $f$ continuous at $x=0$?

Worked Solution: (a) For $x\to 0^{-}$, use the first piece. Direct substitution gives $\frac{0}{0}$, so rationalize by multiplying numerator and denominator by $\sqrt{9+x}+3$: $$\frac{(\sqrt{9+x}-3)(\sqrt{9+x}+3)}{2x(\sqrt{9+x}+3)}=\frac{(9+x)-9}{2x(\sqrt{9+x}+3)}=\frac{x}{2x(\sqrt{9+x}+3)}$$ Cancel $x$ (valid since $x\ne 0$), then substitute $x=0$: $\frac{1}{2(3+3)}=\frac{1}{12}$. So ${\lim }_{x\to 0^{-}}f(x)=\frac{1}{12}$. (b) For ${\lim }_{x\to 0}f(x)$ to exist, ${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{-}}f(x)=\frac{1}{12}$. Evaluate the right-hand limit: ${\lim }_{x\to 0^{+}}\frac{3x+a}{x+2}=\frac{a}{2}$. Set equal to $\frac{1}{12}$: $\frac{a}{2}=\frac{1}{12}⟹a=\frac{1}{6}$. (c) For $f$ to be continuous at $x=0$, we need ${\lim }_{x\to 0}f(x)=f(0)$. We have ${\lim }_{x\to 0}f(x)=\frac{1}{12}$, and $f(0)=\frac{0+a}{0+2}=\frac{a}{2}$. Setting equal gives $a=\frac{1}{6}$, the same value as in part (b).

Question 3 (Application / Real-World Style)

The concentration $C(t)$ (in mg/L) of a drug in a patient's bloodstream $t$ hours after the start of an extended injection is given by $C(t)=\frac{0.4t(t-2)}{t^2+2t-8}$ for $t>0,t\ne 2$. Pharmacologists need to find the limit of the concentration as $t$ approaches 2 hours, when the injection finishes entering the bloodstream. Use algebraic properties of limits to find this limit, and interpret the result.

Worked Solution: We need to evaluate ${\lim }_{t\to 2}\frac{0.4t(t-2)}{t^2+2t-8}$. Direct substitution gives $\frac{0}{0}$, so factor the denominator: $t^2+2t-8=(t+4)(t-2)$. Cancel the common $(t-2)$ factor, which is valid because $t\to 2$, so $t\ne 2$. This simplifies the limit to ${\lim }_{t\to 2}\frac{0.4t}{t+4}$. Substitute $t=2$: $\frac{0.4(2)}{2+4}=\frac{0.8}{6}\approx 0.133$. In context, this means that as the injection finishes at 2 hours, the concentration of the drug in the patient's bloodstream approaches approximately 0.133 mg/L.

8. Quick Reference Cheatsheet

9. What's Next

This topic is the foundational algebraic tool for every subsequent limit-based topic in AP Calculus BC. Next, you will apply these algebraic limit properties to evaluate infinite limits and limits at infinity, which are used to find vertical and horizontal asymptotes of functions, and later to test for convergence of series. Without mastering the algebraic properties of limits introduced here, you will not be able to simplify expressions correctly for these more advanced problems, or for core topics like finding derivatives via the limit definition or evaluating improper integrals. This topic also builds the key intuition that indeterminate forms require further work, which is critical when you learn L'Hospital's Rule for more complicated indeterminate forms later in the course.

• Infinite limits and limits at infinity
• Continuity of functions
• Derivatives from the limit definition
• L'Hospital's Rule