Selecting procedures for determining limits

AP Calculus BC· AP Calculus BC CED — Limits and Continuity· 14 min read

1. Direct Substitution for Determinate Limits★☆☆☆☆⏱ 3 min

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Direct substitution is always the first procedure you should test, as it is the fastest and simplest method. It works when a function $f (x)$ is continuous at $x = a$ , the point we are approaching. If you get a finite real number after substitution, that is your limit and you are done. If you get $c /0$ for $c \neq = 0$ , the limit is infinite or does not exist, which is still a determinate result.

📘 Definition

Direct Substitution

\lim_{x \to a} f(x) = f(a)

If $f$ is continuous at $a$ , the limit equals the function value at $a$ . Applies to all continuous functions at points in their domain.

Example:

Polynomials, rational functions with non-zero denominator at $a$ , trigonometric, exponential, and logarithmic functions.

📐 Worked Example

Evaluate $lim_{x \to 2} \frac{x ^{3} - 4 x + 1}{x ^{2} + 3}$

1
Check if direct substitution is applicable: the function is rational, evaluate the denominator at $x = 2$ :
2
$2^2 + 3 = 7 \neq 0$
3
The function is continuous at $x = 2$ , so substitute into the numerator:
4
$2^3 - 4(2) + 1 = 8 - 8 + 1 = 1$
5
The result is a finite real number, so this is the limit:
6
$\lim_{x \to 2} \frac{x^3 - 4x + 1}{x^2 + 3} = \frac{1}{7}$

Exam tip:

Always test direct substitution first. Roughly 70% of basic AP limit problems can be solved this way, saving valuable exam time.

2. Algebraic Manipulation for Indeterminate $0/0$ Limits★★☆☆☆⏱ 4 min

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When direct substitution gives an indeterminate $0/0$ form, algebraic manipulation is the next procedure to try for problems with polynomials or radicals. If both numerator and denominator are zero at $x = a$ , $(x - a)$ is always a common factor that can be canceled, which does not change the limit since the limit only depends on values near $x = a$ , not at $x = a$ .

📐 Worked Example

Evaluate $lim_{x \to 9} \frac{x - 9}{x - 3}$

1
Test direct substitution: substituting $x = 9$ gives $0/0$ , an indeterminate form requiring algebra.
2
Multiply numerator and denominator by the conjugate of $x - 3$ , which is $x + 3$ :
3
$\lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} = \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}$
4
Cancel the common $(x - 9)$ factor (valid for $x \neq = 9$ , so it does not change the limit):
5
$\lim_{x \to 9} (\sqrt{x} + 3)$
6
Use direct substitution to get the final result:
7
$\sqrt{9} + 3 = 6$

Exam tip:

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

3. L'Hospital's Rule for Indeterminate Forms★★★☆☆⏱ 4 min

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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 hard-to-factor high-degree polynomials. It only applies to indeterminate $0/0$ or $\infty/\infty$ forms, but can be adapted for other indeterminate forms by rewriting them as a fraction.

📘 Definition

L'Hospital's Rule

If $lim_{x \to a} f (x)$ and $lim_{x \to a} g (x)$ are both 0 or both $\pm \infty$ , then $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ provided the right-hand limit exists.

Example:

Indeterminate forms involving transcendental functions.

📐 Worked Example

Evaluate $lim_{x \to 0} \frac{s i n ( 2 x ^{2} )}{l n ( 1 + x ^{2} )}$

1
Test direct substitution: $sin (2 (0)^{2}) = 0$ and $ln (1 + 0) = 0$ , so we have a valid indeterminate $0/0$ form.
2
Differentiate numerator and denominator separately:
3
$f(x) = \sin(2x^2) \implies f'(x) = 4x \cos(2x^2) \\ g(x) = \ln(1+x^2) \implies g'(x) = \frac{2x}{1+x^2}$
4
Rewrite the limit as the ratio of derivatives and simplify:
5
$\lim_{x \to 0} \frac{4x \cos(2x^2)}{\frac{2x}{1+x^2}} = \lim_{x \to 0} 2(1+x^2) \cos(2x^2)$
6
Use direct substitution to get the final result:
7
$2(1+0)\cos(0) = 2(1)(1) = 2$

Exam tip:

Never apply L'Hospital's Rule to determinate forms, and never differentiate the entire quotient with the quotient rule—you must differentiate numerator and denominator separately.

4. Squeeze Theorem for Oscillating Bounded Limits★★★☆☆⏱ 3 min

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The Squeeze Theorem (also called the Sandwich Theorem) is the correct procedure for limits involving bounded oscillating functions, like $sin (\frac{1}{x})$ or $cos (\frac{1}{x})$ , multiplied by a function that approaches zero (or for bounded functions divided by a function approaching infinity). Trigonometric functions like sine and cosine are always bounded between $- 1$ and $1$ , which lets us squeeze the function between two bounds that approach the same limit.

📘 Definition

Squeeze Theorem

If for all $x$ near $a$ (except possibly at $a$ ), $g (x) \leq f (x) \leq h (x)$ , and $lim_{x \to a} g (x) = lim_{x \to a} h (x) = L$ , then $lim_{x \to a} f (x) = L$ .

Example:

Oscillating bounded functions multiplied by terms approaching zero.

📐 Worked Example

Evaluate $lim_{x \to 0} x^{2} cos (\frac{5}{x})$

1
Recognize that $cos (\frac{5}{x})$ is bounded between $- 1$ and $1$ for all $x \neq = 0$ .
2
Write the inequality for the entire function ( $x^{2}$ is non-negative, so inequality signs do not flip):
3
$-x^2 \leq x^2 \cos\left(\frac{5}{x}\right) \leq x^2$
4
Evaluate the limits of the lower and upper bounds:
5
$\lim_{x \to 0} -x^2 = 0 \quad \lim_{x \to 0} x^2 = 0$
6
By the Squeeze Theorem, the limit of the middle function equals 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 correct—algebra and L'Hospital's Rule will not work here.

5. Common Pitfalls

Wrong move:

Applying L'Hospital's Rule to $lim_{x \to 1} \frac{x ^{2} + 1}{x - 1}$ , getting $lim_{x \to 1} \frac{2 x}{1} = 2$ , and concluding the limit is $2$ .

Why:

You 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:

You 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 $x \neq = a$ , so the limit is unchanged, but the original function is still undefined at $x = a$ .

Wrong move:

For $lim_{x \to 0} \frac{e ^{2 x} - 1}{x}$ , after L'Hospital's Rule, incorrectly writing the derivative of the denominator $x$ as $0$ , leading to an undefined answer.

Why:

You focus on differentiating the more complex numerator and overlook the simple denominator.

Correct move:

After writing $f^{'} (x)$ for the numerator, always explicitly write $g^{'} (x)$ for the denominator before simplifying.

Wrong move:

Spending 5 minutes trying to factor $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$ as a polynomial.

Why:

You 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 $lim_{x \to \infty} \frac{s i n x}{x}$ does not exist because $sin x$ oscillates.

Why:

You 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.

6. Quick Reference Cheatsheet

Procedure	When to Use	Key Rule
Direct Substitution	Continuous function at $x = a$ , determinate form	$lim_{x \to a} f (x) = f (a)$ ; stop if you get a finite number
Factoring/Canceling	$0/0$ indeterminate polynomial limits	Cancel common $(x - a)$ factor, then substitute
Conjugate Multiplication	$0/0$ indeterminate limits with radicals	Multiply by conjugate to eliminate radicals, reveal common factors
L'Hospital's Rule	Indeterminate $0/0$ or $\infty/\infty$ , especially transcendental functions	$lim \frac{f}{g} = lim \frac{f ^{'}}{g ^{'}}$ ; differentiate numerator/denominator separately
Squeeze Theorem	Oscillating bounded functions times terms going to 0/ $\infty$	Bound $f (x)$ between two functions with the same limit $L$
Form Check	All limit problems	$0/0, \infty/\infty$ are indeterminate; $c /0 (c \neq = 0)$ is determinate (infinite)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

Most exams · MCQ + FRQ
Standalone and embedded limit problems

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.