Short on time? This page is the recap. For explanations, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: lim (x → a) f(x) means “the limit as x approaches a of f(x)”.
Recap
- Try direct substitution first. If it gives a number with a nonzero denominator, you are done.
- 0/0 is an indeterminate form. It does not tell you the limit. Rewrite the expression.
- Nonzero/0 is not indeterminate: the expression is unbounded near a, so there is no finite limit. Check each side.
- Cancelling a factor (x − a) is valid only for x ≠ a. That is enough, because a limit never uses the value at a.
- Graphically, a cancelled factor shows up as a hole: the limit is the height of the hole.
Key relationships
| Technique | Condition | Key step |
|---|---|---|
| Factor and cancel | Polynomial over polynomial, 0/0 at x = a | Both top and bottom contain (x − a) (factor theorem) |
| Conjugate | Square-root sum or difference gives 0/0 | (p − q)(p + q) = p² − q²; expand only the side with the root |
| Combine fractions | Fraction inside a fraction gives 0/0 | Common denominator; note a − x = −(x − a) |
| Trig identity | Trig expression gives 0/0 | sin²x = (1 − cos x)(1 + cos x); sin 2x = 2 sin x cos x |
| Key fact | f(x) = g(x) near a, x ≠ a | lim (x → a) f(x) = lim (x → a) g(x) |
Useful factorisations: x² − a² = (x − a)(x + a); x³ − a³ = (x − a)(x² + ax + a²).
Assumptions behind the method
- The equivalent expression must agree with the original on an open interval around a (except at a).
- The factor you cancel must be nonzero for x ≠ a near a.
- After simplifying you still substitute, so the simplified expression must be one where substitution is allowed (defined at a).
Mistakes to avoid
- Writing “0/0, so the limit is 0” (or 1, or “does not exist”).
- Cancelling terms, not factors. Only whole factors cancel.
- Dropping “lim” before you substitute.
- Expanding the denominator after multiplying by a conjugate.
- Sign slips when 3 − x appears: 3 − x = −(x − 3).
- Trying to cancel x in sin x / x. That limit is a Topic 1.8 (squeeze theorem) result.
Quick self-check
- Find lim (x → 4) (x² − 16)/(x − 4). (8)
- Find lim (h → 0) ((3 + h)² − 9)/h. (6: the top is 6h + h², so the expression is 6 + h for h ≠ 0)
- Substitution gives 7/0. Can factoring rescue a finite limit? (No. A nonzero number over 0 means the expression is unbounded near a.)
Next: practice questions.