Selecting procedures for determining limits — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers direct substitution, factoring/canceling, conjugate multiplication, the Squeeze Theorem, and L’Hospital’s Rule for indeterminate forms, teaching you to select the correct method for any limit problem on the AP Calculus AB exam.

You should already know: Basic limit laws for sums, products, and quotients. Algebraic manipulation including factoring and rationalizing expressions. The definition of an indeterminate form.

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 Selecting procedures for determining limits?

This topic is the core AP Calculus skill of identifying the structure of a given limit problem and choosing the most efficient, correct solution method, rather than relying on forced application of memorized rules. Per the AP Calculus AB Course and Exam Description (CED), this skill is a key component of Unit 1: Limits and Continuity, which accounts for 10-12% of the total exam score. This skill is tested in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQs require fast, accurate method selection to save time, while FRQs require you to justify your choice of method for full credit. Many students either overcomplicate simple problems by jumping to advanced methods, or get stuck on complex problems by sticking to overly simple algebraic techniques. Mastering this selection process eliminates unnecessary errors, cuts down on exam time, and prepares you for all future calculus topics that depend on limits, from derivatives to integrals.

2. Algebraic Methods for Indeterminate 0/0 Forms

The first step for any limit problem is always to test direct substitution: if a function $f(x)$ is continuous at $x=a$, then ${\lim }_{x\to a}f(x)=f(a)$ by definition of continuity. If direct substitution gives a real number, that is your limit—no further work is needed. If direct substitution gives $\frac{c}{0}$ where $c\ne 0$, the limit is either $\pm \infty$ or does not exist (check one-sided limits to confirm). The only time we need further algebraic simplification is when direct substitution gives the indeterminate form $\frac{0}{0}$, which indicates a hole (not a vertical asymptote) at $x=a$.

For rational functions with polynomial numerators and denominators, factoring and canceling the common $(x-a)$ factor that creates the 0/0 form works. If the 0/0 form comes from a radical expression (square root) in the numerator or denominator, we use conjugate multiplication to rationalize the expression, then cancel the common factor. This is almost always faster than more advanced methods like L'Hospital's Rule for polynomial/radical 0/0 problems.

Worked Example

Problem: Find ${\lim }_{x\to 4}\frac{x^2-5x+4}{\sqrt{x}-2}$

1. First test direct substitution: Plugging $x=4$ gives a numerator of $16-20+4=0$ and a denominator of $\sqrt{4}-2=0$, so we have the indeterminate $\frac{0}{0}$ form requiring simplification.
2. Factor the quadratic numerator: $x^2-5x+4=(x-4)(x-1)$. The denominator is a radical, so we use conjugate multiplication.
3. Multiply numerator and denominator by the conjugate of the denominator, $\sqrt{x}+2$: $$\frac{(x^2-5x+4)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{(x-4)(x-1)(\sqrt{x}+2)}{x-4}$$
4. Cancel the common $(x-4)$ factor (valid for all $x\ne 4$, which is all we need for a limit as $x\to 4$), leaving ${\lim }_{x\to 4}(x-1)(\sqrt{x}+2)$.
5. Use direct substitution on the simplified function: $(4-1)(2+2)=3\times 4=12$. The limit is $12$.

Exam tip: Always test direct substitution first. 80% of basic limit problems on the AP MCQ section can be solved immediately with direct substitution if the function is continuous at the limit point.

3. The Squeeze Theorem for Bounded Oscillating Functions

The Squeeze Theorem is the only valid method for limits involving products of a term approaching zero and a bounded oscillating function (most commonly sine or cosine). The Squeeze Theorem states that if $g(x)\le f(x)\le h(x)$ for all $x$ near $a$ (except possibly $a$ itself), and ${\lim }_{x\to a}g(x)={\lim }_{x\to a}h(x)=L$, then ${\lim }_{x\to a}f(x)=L$.

This method is ideal for functions like $x^n\sin \left(\frac{1}{x}\right)$ or $x\cos x$ as $x\to 0$, because sine and cosine are always bounded between $-1$ and $1$ regardless of their input. We can use this boundedness to "squeeze" the product between two functions that both approach the same limit, forcing the original function's limit to equal that value. No algebraic simplification or L'Hospital's Rule works for these problems, so recognizing when to use the Squeeze Theorem is critical.

Worked Example

Problem: Find ${\lim }_{x\to 0}{x}^2\sin \left(\frac{3}{x}\right)$

1. Direct substitution fails: as $x\to 0$, $\frac{3}{x}$ approaches $\pm \infty$, and $\sin \left(\frac{3}{x}\right)$ oscillates infinitely between $-1$ and $1$, giving an indeterminate $0\times$ (bounded oscillation) that cannot be solved with algebra. We use the Squeeze Theorem.
2. Use the boundedness of the sine function: for all $x\ne 0$, $-1\le \sin \left(\frac{3}{x}\right)\le 1$.
3. Multiply all parts of the inequality by $x^2$, which is always non-negative, so the inequality directions do not change: $$-x^2\le x^2\sin \left(\frac{3}{x}\right)\le x^2$$
4. Calculate the limits of the lower and upper bounds: ${\lim }_{x\to 0}-x^2=0$ and ${\lim }_{x\to 0}{x}^2=0$.
5. By the Squeeze Theorem, the limit of the middle function must also equal $0$.

Exam tip: AP FRQs require you to write the full inequality bound to get full credit for a Squeeze Theorem problem—never just state "the limit is 0" without justifying the inequality.

4. L'Hospital's Rule for Transcendental Indeterminate Forms

L'Hospital's Rule is a powerful method for indeterminate forms of type $\frac{0}{0}$ or $\frac{\infty }{\infty }$, especially when algebraic simplification is difficult or impossible (for example, limits involving exponential, logarithmic, or trigonometric transcendental functions). The rule states that if ${\lim }_{x\to a}f(x)=0$ and ${\lim }_{x\to a}g(x)=0$ (or both approach $\pm \infty$), then: $$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ provided the limit on the right exists (or is $\pm \infty$). L'Hospital's Rule can be applied multiple times in sequence if you still get an indeterminate form after the first application, but it only works for $\frac{0}{0}$ or $\frac{\infty }{\infty }$—it can never be used for other indeterminate forms like $0\times \infty$ without rewriting the product as a quotient first. For method selection, L'Hospital's Rule is almost always faster than algebraic methods for indeterminate forms involving transcendental functions.

Worked Example

Problem: Find ${\lim }_{x\to 0}\frac{e^{2x}-1-2x}{x^2}$

1. Test direct substitution: at $x=0$, the numerator is $e^0-1-0=0$ and the denominator is $0^2=0$, so we have a valid $\frac{0}{0}$ indeterminate form for L'Hospital's Rule.
2. Differentiate the numerator and denominator separately (do not use the quotient rule): $f(x)=e^{2x}-1-2x$ gives $f^{\prime }(x)=2e^{2x}-2$, and $g(x)=x^2$ gives $g^{\prime }(x)=2x$.
3. We now have ${\lim }_{x\to 0}\frac{2e^{2x}-2}{2x}={\lim }_{x\to 0}\frac{e^{2x}-1}{x}$. Testing direct substitution again gives $\frac{0}{0}$, so we can apply L'Hospital's Rule a second time.
4. Differentiate again: new numerator derivative is $2e^{2x}$, new denominator derivative is $1$, giving ${\lim }_{x\to 0}2e^{2x}$.
5. Direct substitution gives $2e^0=2$, so the original limit is $2$.

Exam tip: Always confirm you have a $\frac{0}{0}$ or $\frac{\infty }{\infty }$ form before applying L'Hospital's Rule. AP MCQs often include trap options that come from misapplying the rule to non-indeterminate forms.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Applying L'Hospital's Rule to ${\lim }_{x\to 2}\frac{x+3}{x-2}$ after getting $\frac{5}{0}$, resulting in an incorrect answer of $1$. Why: Students confuse the indeterminate $\frac{0}{0}$ with $\frac{c}{0}$ for $c\ne 0$, and automatically apply L'Hospital's because the denominator is zero. Correct move: Always check the numerator value—non-zero numerator with zero denominator means the limit is infinite (or DNE after checking one-sided limits), no L'Hospital's needed.
• Wrong move: Trying to factor ${\lim }_{x\to 0}{x}^2\cos \left(\frac{1}{x}\right)$ instead of using the Squeeze Theorem, getting stuck and wasting time. Why: Students default to algebraic methods for all limit problems with zero, forgetting that bounded oscillating functions require the Squeeze Theorem. Correct move: If you have a bounded function (sine, cosine) multiplied by a term approaching zero, immediately reach for the Squeeze Theorem.
• Wrong move: Incorrect canceling in ${\lim }_{x\to 0}\frac{x+\sin x}{x}={\lim }_{x\to 0}1+\sin x=1$. Why: Students cancel only the first $x$ in the numerator instead of distributing the division to all terms. Correct move: Always distribute the denominator to every term in the numerator before canceling: $\frac{x+\sin x}{x}=\frac{x}{x}+\frac{\sin x}{x}=1+\frac{\sin x}{x}$.
• Wrong move: Using the quotient rule to differentiate when applying L'Hospital's Rule, getting an incorrect derivative. Why: Students confuse differentiating numerator and denominator separately with differentiating the entire quotient. Correct move: Remember L'Hospital's Rule gives $\frac{f^{\prime }}{g^{\prime }}$, not ${\left(\frac{f}{g}\right)}^{\prime }$, so differentiate top and bottom individually.
• Wrong move: Jumping to L'Hospital's Rule for a simple polynomial 0/0 problem like ${\lim }_{x\to 1}\frac{x^2-1}{x-1}$, making an arithmetic error in differentiation. Why: Students think advanced methods are always better, but simple algebra is less error-prone for polynomial problems. Correct move: Always try factoring first for polynomial 0/0 problems—it's faster and has fewer steps.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

${\lim }_{h\to 0}\frac{(2+h{)}^3-8}{h}$ is equal to which of the following? A) $0$ B) $8$ C) $12$ D) The limit does not exist

Worked Solution: First test direct substitution: plugging $h=0$ gives $\frac{8-8}{0}=\frac{0}{0}$, an indeterminate form. We can solve this with factoring: expand the numerator to get $(8+12h+6h^2+h^3)-8=h(12+6h+h^2)$. Cancel the common $h$ term (valid for $h\ne 0$), then use direct substitution to get ${\lim }_{h\to 0}(12+6h+h^2)=12$. With L'Hospital's Rule, we get the same result: differentiate numerator $3(2+h{)}^2$ and denominator $1$, then substitute $h=0$ to get $3(2{)}^2=12$. The correct answer is C.

Question 2 (Free Response)

Let $f(x)=\frac{x^2-9}{x^2-2x-3}$. (a) Find ${\lim }_{x\to 3}f(x)$, justify your selection of method. (b) Find ${\lim }_{x\to -1}f(x)$, justify your answer. (c) Find ${\lim }_{x\to \infty }f(x)$, using an appropriate method.

Worked Solution: (a) Direct substitution at $x=3$ gives $\frac{9-9}{9-6-3}=\frac{0}{0}$, an indeterminate form, so we use factoring: $x^2-9=(x-3)(x+3)$ and $x^2-2x-3=(x-3)(x+1)$. Cancel the common $(x-3)$ factor to get $\frac{x+3}{x+1}$. Direct substitution gives $\frac{3+3}{3+1}=\frac{6}{4}=\frac{3}{2}$. ${\lim }_{x\to 3}f(x)=\frac{3}{2}$. (b) Direct substitution at $x=-1$ gives $\frac{1-9}{1+2-3}=\frac{-8}{0}$, a non-zero over zero form. Checking one-sided limits: as $x\to -1^{+}$, $f(x)\to +\infty$, and as $x\to -1^{-}$, $f(x)\to -\infty$. Since the one-sided limits do not agree, ${\lim }_{x\to -1}f(x)$ does not exist. (c) For a limit at infinity of a rational function, divide numerator and denominator by the highest power of $x$, $x^2$: $f(x)=\frac{1-\frac{9}{x^2}}{1-\frac{2}{x}-\frac{3}{x^2}}$. As $x\to \infty$, all terms with $\frac{1}{x^n}$ approach $0$, so ${\lim }_{x\to \infty }f(x)=\frac{1}{1}=1$.

Question 3 (Application / Real-World Style)

A coffee shop models the average wait time $W(q)$ in minutes for a customer as a function of the number of baristas working, $q$, given by $W(q)=\frac{12q+8}{2q+1}$ for $q\ge 1$. Find the limiting average wait time as the shop schedules more and more baristas (${\lim }_{q\to \infty }W(q)$), and interpret your result in context.

Worked Solution: This is a limit at infinity of a rational function. Divide numerator and denominator by the highest power of $q$, which is $q$: $$W(q)=\frac{12+\frac{8}{q}}{2+\frac{1}{q}}$$ As $q\to \infty$, $\frac{8}{q}\to 0$ and $\frac{1}{q}\to 0$, so ${\lim }_{q\to \infty }W(q)=\frac{12}{2}=6$ minutes. In context, this means that even with an extremely large number of baristas working, the minimum average wait time will approach 6 minutes due to order processing time per customer, even when there is no wait for a free barista.

7. Quick Reference Cheatsheet

8. What's Next

Mastering limit procedure selection is the foundational prerequisite for all of calculus, starting with the definition of the derivative, which is itself a limit of a difference quotient. Every derivative shortcut you use comes from simplifying a limit to derive the rule, so if you cannot select the right method for limits, you will struggle to connect the formal definition of the derivative to practical calculation. Next, you will apply your limit skills to define continuity at a point, identify different types of discontinuities, and analyze the end behavior of functions. Later in the course, limits are used to evaluate improper integrals and calculate areas under infinite curves, which are tested on the AP exam.

• Defining continuity and discontinuity
• Formal definition of the derivative
• Evaluating improper integrals