Selecting procedures for determining limits — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers direct substitution, factoring/canceling, conjugate multiplication, L'Hospital's Rule, and the Squeeze Theorem for selecting the correct procedure to evaluate one-sided and two-sided limits of algebraic, trigonometric, and transcendental functions.

You should already know: Basic limit notation, algebraic manipulation including factoring and conjugates, and properties of continuous functions.

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 Selecting procedures for determining limits?

According to the AP Calculus BC Course and Exam Description (CED), this topic is part of Unit 1: Limits and Continuity, which accounts for 10–12% of the total AP exam score. Selecting procedures for determining limits is the core decision-making skill behind every limit problem: instead of just memorizing rules, you learn to quickly identify which technique is appropriate for a given limit, saving time and avoiding common errors on both multiple-choice (MCQ) and free-response (FRQ) sections. This skill appears across the exam: it is tested directly in standalone MCQs, required as an early step in derivative and integral problems, and often embedded in the first part of multi-part FRQs. Unlike rote calculation, this topic emphasizes pattern recognition: you will learn to categorize limit problems by their form (indeterminate vs. determinate) and apply the matching strategy. Many students rush into applying advanced techniques like L'Hospital's Rule to problems that can be solved much faster with basic algebra, leading to unnecessary mistakes; this chapter teaches you to prioritize efficient, correct procedures.

2. Direct Substitution for Determinate Limits

Direct substitution is always the first procedure you should test, because it is the fastest and simplest method. Direct substitution works when a function $f(x)$ is continuous at the point $x=a$ that we are approaching. By the definition of continuity, if $f$ is continuous at $a$, then ${\lim }_{x\to a}f(x)=f(a)$. This is applicable to all polynomials, rational functions where the denominator at $a$ is non-zero, and all trigonometric, exponential, and logarithmic functions at points in their domain. If you substitute $x=a$ and get a finite real number, that is your limit—you are done. If you get a non-zero number over zero (e.g., $5/0$), that means the limit is either $+\infty$, $-\infty$, or does not exist (confirm by checking one-sided limits), and this is still a determinate form, not indeterminate. Only when you get $0/0$ or $\infty /\infty$ (or other indeterminate forms like $0\cdot \infty$, $\infty -\infty$) do you need to move to another procedure.

Worked Example

Evaluate ${\lim }_{x\to 2}\frac{x^3-4x+1}{x^2+3}$.

1. Check if direct substitution is applicable: The function is rational, so evaluate the denominator at $x=2$: $2^2+3=7\ne 0$, so the function is continuous at $x=2$.
2. Substitute $x=2$ into the numerator: $2^3-4(2)+1=8-8+1=1$.
3. The result is $\frac{1}{7}$, a finite real number, so this is the limit.
4. Final answer: ${\lim }_{x\to 2}\frac{x^3-4x+1}{x^2+3}=\frac{1}{7}$

Exam tip: Always test direct substitution first. Roughly 70% of basic limit problems on the AP exam can be solved with direct substitution, so you will save valuable exam time if you check this before reaching for more complex techniques.

3. Algebraic Manipulation for Indeterminate $0/0$ Limits

When direct substitution gives an indeterminate $0/0$ form, algebraic manipulation is the next procedure to try for problems involving polynomials or radicals. Two common algebraic methods are factoring/canceling and conjugate multiplication. Factoring works for polynomials: if both numerator and denominator evaluate to zero at $x=a$, then $(x-a)$ must be a common factor of both. We can factor out and cancel this common factor, which does not change the limit (since the limit only depends on values near $x=a$, not at $x=a$), then evaluate the resulting function with direct substitution. Conjugate multiplication works when one term in the numerator or denominator is a difference of radicals: multiplying by the conjugate eliminates the radical, revealing a common factor that can be canceled.

Worked Example

Evaluate ${\lim }_{x\to 9}\frac{x-9}{\sqrt{x}-3}$.

1. Test direct substitution: Substitute $x=9$, numerator $9-9=0$, denominator $\sqrt{9}-3=0$, so we have an indeterminate $0/0$ form requiring algebraic manipulation.
2. Multiply numerator and denominator by the conjugate of $\sqrt{x}-3$, which is $\sqrt{x}+3$: $$\underset{x\to 9}{\lim }\frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\underset{x\to 9}{\lim }\frac{(x-9)(\sqrt{x}+3)}{x-9}$$
3. Cancel the common $(x-9)$ factor, which is valid for all $x\ne 9$ so it does not change the limit, leaving ${\lim }_{x\to 9}(\sqrt{x}+3)$.
4. Use direct substitution on the simplified expression: $\sqrt{9}+3=6$, so the limit is $6$.

Exam tip: After canceling a common factor, always re-test direct substitution on the simplified expression. For $0/0$ polynomial/radical limits, you will almost always get a finite answer after simplification.

4. L'Hospital's Rule for Indeterminate Forms

L'Hospital's Rule is the go-to procedure for indeterminate forms when algebraic manipulation is not feasible, such as for problems involving transcendental functions (trigonometric, exponential, logarithmic) or high-degree polynomials that are difficult to factor. L'Hospital's Rule states that if ${\lim }_{x\to a}f(x)=0$ and ${\lim }_{x\to a}g(x)=0$, OR ${\lim }_{x\to a}f(x)=\pm \infty$ and ${\lim }_{x\to a}g(x)=\pm \infty$, 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 $\pm \infty$). This rule only applies to indeterminate forms $0/0$ or $\infty /\infty$, but it can be adapted for other indeterminate forms: $0\cdot \infty$ can be rewritten as a fraction to get one of the two valid forms, and $\infty -\infty$ can be combined into a single fraction to get a valid indeterminate form.

Worked Example

Evaluate ${\lim }_{x\to 0}\frac{\sin (2x^2)}{\ln (1+x^2)}$.

1. Test direct substitution: $\sin (2(0{)}^2)=0$ and $\ln (1+0)=0$, so we have a valid indeterminate $0/0$ form for L'Hospital's Rule.
2. Differentiate numerator and denominator separately: $f(x)=\sin (2x^2)⟹f^{\prime }(x)=4x\cos (2x^2)$, and $g(x)=\ln (1+x^2)⟹g^{\prime }(x)=\frac{2x}{1+x^2}$.
3. Rewrite the limit as the ratio of derivatives and simplify: $$\underset{x\to 0}{\lim }\frac{4x\cos (2x^2)}{\frac{2x}{1+x^2}}=\underset{x\to 0}{\lim }2(1+x^2)\cos (2x^2)$$
4. Use direct substitution: $2(1+0)\cos (0)=2(1)(1)=2$, so the limit is $2$.

Exam tip: Never apply L'Hospital's Rule to determinate forms, and never differentiate the entire quotient with the quotient rule—you must differentiate the numerator and denominator separately, which is one of the most common student mistakes on the exam.

5. Squeeze Theorem for Oscillating Bounded Limits

The Squeeze Theorem (also called the Sandwich Theorem) is the correct procedure for limits involving bounded oscillating functions, like $\sin \left(\frac{1}{x}\right)$ or $\cos \left(\frac{1}{x}\right)$, multiplied by a function that approaches zero. The Squeeze Theorem states that if for all $x$ near $a$ (except possibly at $a$), we have $g(x)\le f(x)\le h(x)$, and ${\lim }_{x\to a}g(x)={\lim }_{x\to a}h(x)=L$, then ${\lim }_{x\to a}f(x)=L$. Trigonometric functions like sine and cosine are always bounded between $-1$ and $1$, so we can use this bound to squeeze the entire product between two functions that both approach the same limit.

Worked Example

Evaluate ${\lim }_{x\to 0}{x}^2\cos \left(\frac{5}{x}\right)$.

1. Recognize that $\cos \left(\frac{5}{x}\right)$ is bounded between $-1$ and $1$ for all $x\ne 0$, which is all we need for the limit as $x\to 0$.
2. Write the inequality for the entire function: $-x^2\le x^2\cos \left(\frac{5}{x}\right)\le x^2$ (since $x^2$ is non-negative, we do not need to add absolute value or flip inequality signs).
3. Evaluate the limits of the lower and upper bounds: ${\lim }_{x\to 0}-x^2=0$ and ${\lim }_{x\to 0}{x}^2=0$.
4. By the Squeeze Theorem, the limit of the middle function must equal $0$.

Exam tip: If you see a trigonometric function with $\frac{1}{x}$ or another term that causes oscillation as $x\to 0$ or $x\to \infty$, the Squeeze Theorem is almost always the correct procedure—algebraic manipulation and L'Hospital's Rule will not work here.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Applying L'Hospital's Rule to ${\lim }_{x\to 1}\frac{x^2+1}{x-1}$, getting ${\lim }_{x\to 1}\frac{2x}{1}=2$, and concluding the limit is $2$. Why: The student jumped to L'Hospital's Rule without checking the form after direct substitution; the original limit gives $2/0$, a determinate infinite form, not indeterminate. Correct move: Always test direct substitution first, and confirm you have an indeterminate form before applying L'Hospital's Rule.
• Wrong move: For ${\lim }_{x\to 2}\frac{(x-2)(x+3)}{(x-2)(x-5)}$, canceling $(x-2)$ and concluding the function is equal to $\frac{x+3}{x-5}$ everywhere, including at $x=2$. Why: Students confuse the value of the function at $x=2$ with the limit as $x\to 2$, leading to incorrect conclusions about continuity. Correct move: Remember that canceling $(x-a)$ only removes the common factor for $x\ne a$, so the limit is unchanged, but the original function is still undefined at $x=a$.
• Wrong move: For ${\lim }_{x\to 0}\frac{e^{2x}-1}{x}$, after L'Hospital's Rule, incorrectly writing the derivative of the denominator $x$ as $0$, leading to an undefined answer. Why: Students focus on differentiating the more complex numerator and overlook the simple denominator. Correct move: After writing $f^{\prime }(x)$ for the numerator, always explicitly write $g^{\prime }(x)$ for the denominator before simplifying.
• Wrong move: Spending 5 minutes trying to factor ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$ as a polynomial. Why: Students default to algebraic manipulation for any $0/0$ limit, regardless of function type. Correct move: If you have a $0/0$ limit with transcendental functions, reach for L'Hospital's Rule or standard trigonometric limits immediately.
• Wrong move: Concluding ${\lim }_{x\to \infty }\frac{\sin x}{x}$ does not exist because $\sin x$ oscillates. Why: Students forget the Squeeze Theorem applies to limits at infinity as well as finite points. Correct move: If you have a bounded function divided by a function going to infinity, set up the Squeeze Theorem inequality to find the limit.

7. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

${\lim }_{x\to 4}\frac{3-\sqrt{5+x}}{x^2-16}$ is equal to which of the following? A) $-\frac{1}{48}$ B) $-\frac{1}{12}$ C) $\frac{1}{48}$ D) The limit does not exist

Worked Solution: First test direct substitution: substituting $x=4$ gives $0/0$, an indeterminate form. We use conjugate multiplication: multiply numerator and denominator by $3+\sqrt{5+x}$ to get $\frac{9-(5+x)}{(x^2-16)(3+\sqrt{5+x})}=\frac{-(x-4)}{(x-4)(x+4)(3+\sqrt{5+x})}$. Cancel the $(x-4)$ factor, then substitute $x=4$ to get $\frac{-1}{(8)(6)}=-\frac{1}{48}$. The correct answer is A.

Question 2 (Free Response)

Consider the function $f(x)=\frac{x^2-3x+2}{\sin (\pi x)}$. (a) Evaluate ${\lim }_{x\to 1}f(x)$. (b) Explain why direct substitution fails to give the limit in (a), and what procedure you used to get your answer. (c) Evaluate ${\lim }_{x\to \infty }\frac{\ln (2x)}{e^{x/2}}$.

Worked Solution: (a) Direct substitution at $x=1$ gives $\frac{1-3+2}{\sin (\pi )}=\frac{0}{0}$, an indeterminate form. Apply L'Hospital's Rule: differentiate numerator to get $2x-3$, differentiate denominator to get $\pi \cos (\pi x)$. Substitute $x=1$: $\frac{2(1)-3}{\pi \cos (\pi )}=\frac{-1}{\pi (-1)}=\frac{1}{\pi }$. Final answer: $\boxed{\frac{1}{\pi }}$. (b) Direct substitution fails because substituting $x=1$ gives the indeterminate form $0/0$, which is not a finite value and does not indicate an infinite limit. We used L'Hospital's Rule here because we have an indeterminate form involving a transcendental function, which is faster than factoring (though factoring the numerator gives the same result). (c) As $x\to \infty$, $\ln (2x)\to \infty$ and $e^{x/2}\to \infty$, so we have indeterminate $\infty /\infty$. Apply L'Hospital's Rule: derivative of numerator is $\frac{1}{x}$, derivative of denominator is $\frac{1}{2}{e}^{x/2}$. The limit becomes ${\lim }_{x\to \infty }\frac{2}{x{e}^{x/2}}=0$. Final answer: $\boxed{0}$.

Question 3 (Application / Real-World Style)

In population biology, the per-capita growth rate of a population with size $N$ is given by $r(N)=\frac{r_{0}K\ln \left(\frac{K}{N}\right)}{K-N}$, where $r_0=0.2$ per year is the intrinsic growth rate, and $K=1000$ individuals is the carrying capacity of the environment. Find ${\lim }_{N\to K}r(N)$, and interpret what this limit means in context.

Worked Solution: Substitute $N=K$ to get $\frac{0}{0}$, an indeterminate form. Apply L'Hospital's Rule, with $f(N)=r_{0}K(\ln K-\ln N)$ and $g(N)=K-N$. Compute derivatives: $f^{\prime }(N)=-\frac{r_{0}K}{N}$ and $g^{\prime }(N)=-1$. The limit becomes ${\lim }_{N\to K}\frac{-r_{0}K/N}{-1}=\frac{r_{0}K}{K}=r_0=0.2$. Final answer: The limit is 0.2 per year. Interpretation: As the population size approaches the environment's carrying capacity, the per-capita growth rate approaches the intrinsic growth rate of 0.2 individuals per individual per year.

8. Quick Reference Cheatsheet

9. What's Next

Mastering selection of limit procedures is the foundational prerequisite for all upcoming topics in AP Calculus BC. Immediately after this unit, you will use limit evaluation to define the derivative via the difference quotient, and later to define definite integrals as limits of Riemann sums. Without the ability to quickly select and apply the correct limit procedure, you will struggle to compute derivative definitions and improper integrals, which both rely on core limit skills. This topic also feeds into more advanced topics like series convergence, where you will apply limit comparison tests and ratio tests that require evaluating limits of sequences. Continue your study with these related topics: Defining limits and using limit notation Continuity of functions Defining the derivative as a limit Limits of sequences for infinite series