Calculus I › Applications of derivatives › full formula sheet
Say it: “for a zero-over-zero or infinity-over-infinity form, the limit of f over g equals the limit of f prime over g prime”
L’Hôpital’s rule
When plugging in gives 0/0, differentiate the top and bottom instead.
Named for Guillaume de l’Hôpital (1696) — though Johann Bernoulli discovered it. f′/g′ means differentiate f and g separately: this is not the quotient rule.
Before this lesson: Definition of the derivative · Mean Value Theorem
Where it comes from
The problem: limx→0 sin x / x. Plugging in gives 0/0 — and the naive readings all fail:
The right intuition is linearization: near a, with f(a) = g(a) = 0,
Both functions vanish at a, so near a each is nearly its tangent line through (a, 0) — and the (x−a)’s cancel, leaving the ratio of slopes. For sin x/x at 0: cos 0 / 1 = 1. ✓
Before reading on: near x = a, f(x) ≈ f′(a)(x−a) and g(x) ≈ g′(a)(x−a). What happens to the ratio f(x)/g(x) — and what cancels?
Derivation
The linearization intuition, made rigorous by Cauchy’s Mean Value Theorem: for f, g continuous on [a, x] and differentiable on (a, x), some c between a and x has [f(x)−f(a)]/[g(x)−g(a)] = f′(c)/g′(c).
Note the fine print: the proof needs lim f′/g′ to exist (and g′ ≠ 0 near a). When it doesn’t — like (x + sin x)/x as x→∞, where f′/g′ = 1 + cos x oscillates — the rule simply doesn’t apply. The original limit may still exist (it’s 1, by squeeze).
Before reading on: limx→0 (x²+1)/x: can you apply L’Hôpital? What’s the form — and what goes wrong if you differentiate anyway?
How to use it
The procedure, every time:
- Check the form FIRST. Plug in: you must get 0/0 or ∞/∞. Anything else — stop. Applying L’Hôpital to a non-indeterminate form manufactures wrong answers.
- Differentiate the top and bottom separately. f′/g′ — this is not the quotient rule. Say it once: “top-prime over bottom-prime.”
- Take the limit again. If it’s still 0/0 or ∞/∞, repeat from Step 1.
- Convert other forms first: 0·∞ → rewrite as a quotient (e.g. x·ln x = ln x / (1/x)); ∞−∞ → combine into one fraction; 1∞, 00, ∞0 → take logarithms.
When to reach for it
Any 0/0 or ∞/∞ limit where algebra (factoring, conjugates, known limits) stalls. It’s the power tool — but check the form first, every time.
When it fails
If lim f′/g′ doesn’t exist, the rule says nothing — try squeeze, algebra, or known limits instead. Failure of the rule ≠ failure of the limit.
Worked examples
Four limits, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the classic: limx→0 sin x / x
- Check the form. sin 0 / 0 = 0/0. ✓ — indeterminate, rule applies.
- Differentiate top and bottom separately. (sin x)′ = cos x; (x)′ = 1. New limit: limx→0 cos x / 1.
- Evaluate. cos 0 / 1 = 1.
Your turn: limx→0 tan x / x.
Answer: 1.
0/0 ✓; (tan x)′ = sec²x, (x)′ = 1 — new limit sec²0 = 1.
Example 2 — exponentials: limx→0 (ex − 1)/x
- Check the form. (1 − 1)/0 = 0/0. ✓
- Differentiate separately. (ex − 1)′ = ex; (x)′ = 1. New limit: limx→0 ex/1.
- Evaluate. e0 = 1. (This limit is secretly the definition of the derivative of ex at 0.)
Your turn: limx→0 (e2x − 1)/x.
Answer: 2.
0/0 ✓; top′ = 2e2x, bottom′ = 1 — new limit 2e0 = 2.
Example 3 — at infinity: limx→∞ ln x / x
- Check the form. ∞/∞. ✓ — the rule works at ∞ too.
- Differentiate separately. (ln x)′ = 1/x; (x)′ = 1. New limit: limx→∞ (1/x)/1.
- Evaluate. 1/x → 0. (Logs grow slower than lines — L’Hôpital makes it rigorous.)
Your turn: limx→∞ (ln x)/x².
Answer: 0.
∞/∞ ✓; (ln x)′ = 1/x, (x²)′ = 2x — new limit (1/x)/(2x) = 1/(2x²) → 0.
Example 4 — twice: limx→0 (1 − cos x)/x²
- Check the form. (1 − 1)/0 = 0/0. ✓
- First application. (1 − cos x)′ = sin x; (x²)′ = 2x. New limit: limx→0 sin x/(2x) — still 0/0.
- Second application. (sin x)′ = cos x; (2x)′ = 2. New limit: limx→0 cos x/2.
- Evaluate. 1/2. So 1 − cos x ≈ x²/2 near 0 — the parabola the cosine hugs. Answer: 1/2.
Your turn: limx→0 (x − sin x)/x³.
Answer: 1/6.
0/0 → (1 − cos x)/(3x²), still 0/0 → (sin x)/(6x), still 0/0 → (cos x)/6 → 1/6.
Memorization tips
- Form first: plug in — 0/0 or ∞/∞ or walk away. Tattoo this order on the procedure.
- Top-prime over bottom-prime: two separate derivatives. If you see a minus sign, that’s the quotient rule — wrong tool.
- The cancel picture: near a, both functions are their tangent lines through (a, 0); the (x−a)’s cancel, leaving the slope ratio. That image is the rule.
- Repeat as needed: still 0/0? Differentiate again. Example 4 needed two rounds.
- Convert, then conquer: 0·∞ → quotient; ∞−∞ → one fraction; weird powers → logarithms.
- Rule failure ≠ limit failure: (x + sin x)/x as x→∞ breaks the rule (1 + cos x oscillates) but the limit is 1. Have a backup plan.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and L’Hôpital’s rule is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is L’Hôpital’s rule?
If lim f(x)/g(x) has the indeterminate form 0/0 or ∞/∞, then it equals lim f′(x)/g′(x) — differentiate the top and bottom separately (not the quotient rule), then take the limit again.
Why isn’t 0/0 just 0?
0/0 is indeterminate — it can be anything. limx→0 sin x/x = 1, limx→0 x²/x = 0, limx→0 x/x² = ∞. The 0/0 form alone tells you nothing; L’Hôpital’s rule resolves it via the derivatives.
Why does differentiating top and bottom work?
Near a, f(x) ≈ f′(a)(x−a) and g(x) ≈ g′(a)(x−a) (both vanish at a). The (x−a) factors cancel, leaving f′(a)/g′(a). Cauchy’s Mean Value Theorem makes this rigorous.
What must you check before using L’Hôpital’s rule?
That the limit really is 0/0 or ∞/∞ — plug in first. Applying it to a non-indeterminate form gives wrong answers.
When does L’Hôpital’s rule fail?
When lim f′/g′ doesn’t exist, e.g. limx→∞ (x+sin x)/x: differentiating gives 1+cos x, which oscillates — yet the original limit is 1 (squeeze it). Failure of the rule is not failure of the limit.
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