AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/1/1-7)
Unit 1 · Topic 1.7
1.7 Selecting Procedures for Determining Limits
Finding a limit is often more about picking the right method than doing hard algebra. Start with direct substitution, then let the result and the shape of the expression tell you what to try next.
Key terms
- direct substitution
- indeterminate form
- equivalent expressions
- choosing a strategy
Step 1: try direct substitution
Always plug in x = c first. The result sorts the problem into one of three cases:
| Result of substituting | What it means | Next move |
|---|---|---|
| A real number, like 7 or −1/2 | That is the limit (if f is continuous at c) | Done |
| Nonzero number / 0, like 5/0 | Outputs grow without bound | Check signs on each side (1.14) |
| 0/0 | Indeterminate | Rewrite the expression (1.6) |
Step 2: if you get 0/0, read the expression
The look of the expression usually points to the tool. Match it to one of these:
- Polynomials on top and bottom: factor and cancel.
- A square root next to a number, like √(x + 1) − 2: multiply by the conjugate.
- Fractions inside a fraction: combine with a common denominator.
- Trig functions: look for a Pythagorean identity, or a special limit like sin(kx)/x.
- Absolute value: split it into its two pieces and find one-sided limits.
- Piecewise function at a break point: find the left and right limits using the matching pieces.
The special limit sin(kx)/x
Since lim (u→0) sin u / u = 1, you can handle sin(5x) / (3x) by multiplying top and bottom by 5: (5/3)·sin(5x)/(5x). As x→0, 5x→0, so the limit is 5/3 × 1 = 5/3. In general, lim (x→0) sin(ax)/(bx) = a/b.
Equivalent expressions
Every rewriting step must produce an expression that equals the original for all x near c (except possibly c). Factoring, canceling a nonzero factor, multiplying by a form of 1 (like the conjugate over itself) and using identities all keep the expressions equivalent. Squaring both sides or dropping a term does not.
Step 3: decide what kind of answer you have
Once the algebra is done, a limit can end in three ways: a real number, ∞ or −∞ (unbounded in one direction), or does not exist (the sides disagree or the outputs oscillate). Say which one you have, in words or with correct notation.
If the expression has different formulas on each side of c, as with absolute value or piecewise functions, always find both one-sided limits before you write a final answer. A two-sided limit exists only if the two sides agree.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Pick the tool: trig identity
Find lim (x→0) sin²x / (1 − cos x).
Show the solutionHide the solution
- Step 1: Substitute: 0 / (1 − 1) = 0/0, so rewrite.
- Step 2: sin²x = 1 − cos²x = (1 − cos x)(1 + cos x).
- Step 3: Cancel (1 − cos x), which is nonzero for x near 0 but not at 0: the expression equals 1 + cos x.
- Step 4: Substitute: 1 + cos 0 = 2.
Answer: 2
- Example 2
Trap: an absolute value hides two different sides
Find lim (x→2) (x − 2) / |x − 2|.
Show the solutionHide the solution
- Step 1: Substitute: 0/0. The absolute value is the clue, so split by side.
- Step 2: For x > 2, |x − 2| = x − 2, so the expression is 1. The right-hand limit is 1.
- Step 3: For x < 2, |x − 2| = −(x − 2), so the expression is −1. The left-hand limit is −1.
- Step 4: The one-sided limits differ.
Answer: The limit does not exist (left-hand limit −1, right-hand limit 1).
- Example 3
Pick the tool: sum of cubes and a special trig limit
Find (a) lim (x→−1) (x³ + 1) / (x + 1) and (b) lim (x→0) sin(5x) / (3x).
Show the solutionHide the solution
- Step 1: (a) Substitute: 0/0. Use a³ + b³ = (a + b)(a² − ab + b²): x³ + 1 = (x + 1)(x² − x + 1).
- Step 2: Cancel (x + 1) and substitute: (−1)² − (−1) + 1 = 3.
- Step 3: (b) Substitute: 0/0. Rewrite as (5/3)·[sin(5x) / (5x)].
- Step 4: As x→0, 5x→0, so sin(5x)/(5x)→1. The limit is 5/3.
Answer: (a) 3, (b) 5/3
Common mistakes
- Skipping direct substitution and diving into algebra on a limit that was just a plug-in.
- Treating sin(5x)/(3x) as 1 or as 5x/3x without adjusting. The angle inside sine and the bottom must match before the special limit applies.
- Forgetting the minus sign when |x − 2| is rewritten for x < 2.
On the exam
- In multiple choice, several answer options are often the results of common wrong moves. Choose your method first, then work carefully.
- If you're stuck on a 0/0, ask whether the limit is the definition of a derivative (2.2) or whether L'Hospital's Rule (4.7) applies.
Connected topics
Videos
Check yourself
4 questions on 1.7 Selecting Procedures for Determining Limits. Pick an answer to see if you got it, and why.
When x = 1 is substituted into (x² + 3)/(x − 1), the result is 4/0. Which of the following is true about lim (x→1) (x² + 3)/(x − 1) ?
What is lim (x→0) (x² + 2x)/sin x ?
What is lim (x→3) |x − 3|/(x − 3) ?
What is lim (x→−1) (x² + 1)/(x + 1)² ?
0 of 4 answered