Short on time? This page is the recap. For the tangent-line explanation, the graph and three 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
- 0/0 and ∞/∞ are indeterminate forms. They are labels for a type of limit, not values. Never write “= 0/0”.
- The rule: if lim f(x) = 0 and lim g(x) = 0 (or both are infinite), then lim f(x)/g(x) = lim f′(x)/g′(x), as long as the new limit exists or is infinite.
- It works for x → a, for one-sided limits and for x → ±∞.
- Why it works (0/0): near a, each function is close to its tangent line, so the ratio of the functions is close to the ratio of the slopes.
- If the new limit is still 0/0 or ∞/∞, check again and apply the rule again.
Key relationships
| Situation | What to do |
|---|---|
| Substitution gives a number with a nonzero bottom | Done. The rule is not needed (and gives wrong answers if used). |
| Both limits 0, or both infinite | Write the check, then use lim f′(x)/g′(x). |
| Nonzero over 0 | Not indeterminate: unbounded, no finite limit (Topic 1.14). |
| 0 over ∞ | Not indeterminate: the limit is 0. |
| Rule keeps returning the same problem | Use algebra instead, e.g. divide by the highest power. |
| Only f(a), g(a), f′(a), g′(a) given, derivatives continuous, g′(a) ≠ 0 | Limit = f′(a)/g′(a), if f(a) = g(a) = 0. |
The written check (copy this shape): “lim (x → a) f(x) = 0 and lim (x → a) g(x) = 0, so L’Hospital’s Rule applies.”
Assumptions behind the rule
- f and g are differentiable near a (except possibly at a), and g′(x) ≠ 0 there.
- The new limit lim f′(x)/g′(x) exists or is infinite. If it does not exist, the rule gives no conclusion.
- Other forms (0 · ∞, ∞ − ∞, 1^∞ and so on) are not assessed on the exam.
Mistakes to avoid
- Skipping the check, or checking only the first time when the rule is used twice.
- Using the quotient rule on f/g instead of differentiating top and bottom separately.
- Writing “lim f(x)/g(x) = 0/0” as if it were a value.
- Forgetting the chain rule, e.g. d/dx (sin 5x) = 5 cos 5x.
- Using the rule on a non-indeterminate limit such as (x² + 3)/(x + 1) at 0.
- Dropping “lim” before the substitution step.
Quick self-check
- Find lim (x → 0) (sin 4x)/x. (4: the form is 0/0, and the rule gives 4 cos 4x/1 → 4)
- Find lim (x → ∞) x³/eˣ. (0: ∞/∞ three times in a row, ending with 6/eˣ → 0)
- Can you use the rule on lim (x → 0⁺) (cos x)/x? (No. The top tends to 1, not 0. This is 1/0, so the expression is unbounded: it tends to +∞.)
Next: practice questions.