L'Hopital's Rule Applications
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Direct answer
Direct substitution that yields 0/0 or ∞/∞ invites L'Hôpital's rule: differentiate numerator and denominator separately (never the quotient) and re-substitute, repeating while the indeterminate form persists — lim f(x)/g(x) = lim f'(x)/g'(x). The other indeterminate shapes must first be converted: 0·∞ by inverting one factor, ∞ − ∞ by combining into a single fraction, and the power forms 1^∞, 0⁰ and ∞⁰ via A = e^(ln A), which converts the exponent problem into a 0·∞ product. Thus lim (1 + x)^(1/x) as x → 0 is e, because the exponent limit is lim ln(1 + x)/x = 1. Series expansion is the faster sibling: tan x − x over x − sin x tends to 2 by comparing cubic coefficients without any differentiation.
What you must remember
- Applicable forms: only 0/0 and ∞/∞, with both numerator and denominator differentiable near the point and the denominator's derivative non-zero; the rule gives lim f'/g' when that limit exists.
- Differentiate separately: the rule is not the quotient rule — d/dx[f/g] with g² below is a different (wrong) object.
- Repeat with care: apply as many times as the form persists, but stop the moment substitution resolves; differentiating past resolution gives wrong answers.
- Power forms: 1^∞, 0⁰, ∞⁰ → write e^(ln base × exponent) and evaluate the exponent's limit; the e^ln move is mandatory, optional nowhere.
- ∞ − ∞ and 0·∞: combine fractions or invert a factor to reach 0/0 or ∞/∞; rationalise surd differences first (√(x + 1) − √x type).
- Standard limits to cite instead: lim sin x/x = 1, lim (1 + x)^(1/x) = e, lim (e^x − 1)/x = 1, lim ln(1 + x)/x = 1 — using these directly is faster than re-deriving via L'Hôpital.
- Expansion comparison: for small x, sin x ≈ x − x³/6, tan x ≈ x + x³/3, e^x ≈ 1 + x + x²/2; matching leading orders often beats repeated differentiation.
Three limits that teach the method
First, a clean 0/0: lim (e^x − 1 − x)/x² as x → 0. Substituting gives 0/0; differentiate top and bottom: (e^x − 1)/2x, still 0/0; differentiate again: e^x/2 → 1/2. Two honest rounds, each justified by re-checking the form. The expansion route confirms it instantly: e^x − 1 − x ≈ x²/2, so the ratio tends to 1/2.
Second, the exponent trap: lim (tan x/x)^(1/x²) as x → 0. The base tends to 1 and the exponent to ∞ — a 1^∞ form. Take logs: the exponent limit is lim [ln(tan x/x)]/x². Now ln(tan x/x) = ln(1 + (tan x − x)/x) ≈ (tan x − x)/x ≈ (x³/3)/x = x²/3, so the logged limit is 1/3 and the answer is e^(1/3). Without the e^ln conversion there is nothing to differentiate — the form is not a quotient.
Third, a 0·∞ case: lim x·ln x as x → 0⁺ rewrites as ln x/(1/x) → −∞/∞, and one differentiation gives (1/x)/(−1/x²) = −x → 0⁺ — inverting the power rather than the logarithm keeps the algebra human.
When L'Hopital fails
The rule requires the differentiated limit to exist: lim (x + sin x)/x as x → ∞ is a genuine ∞/∞, but differentiating gives (1 + cos x)/1, which oscillates — the rule is silent, yet squeeze thinking (sin x between −1 and 1) gives the limit 1. Students who conclude "limit does not exist" from a failed L'Hôpital round have confused the tool with the truth. The second failure mode is cycling: some quotients reproduce themselves under differentiation; the fix is expansion, not a third application. Third, using the rule where the form is not indeterminate — lim (x + 1)/(x + 2) at x → 1 is 2/3 by substitution; differentiating first gives 1/1 = 1, confidently wrong. Main tests mechanics on clean 0/0 cases; Advanced tests judgment — routing each limit through the standard list, expansion, or the rule — and that routing is the real syllabus of this page.
Frequently asked questions
Which indeterminate forms accept L'Hôpital's rule directly?
Only 0/0 and ∞/∞; every other form (0·∞, ∞ − ∞, 1^∞, 0⁰, ∞⁰) must first be algebraically converted to one of these.
How do you handle 1^∞ limits?
Write the expression as e^(ln base × exponent) and evaluate the exponent's limit, usually a 0·∞ that becomes 0/0 — the classic lim (1 + x)^(1/x) = e works this way.
Why is differentiating numerator and denominator not the quotient rule?
L'Hôpital takes f'(x)/g'(x) as a new fraction; the quotient rule would give [f'g − fg']/g², an entirely different quantity — the rule is not differentiation of the original quotient.
What should you do when L'Hôpital's rule keeps cycling?
Stop and switch tools: series expansion, standard limits, or algebraic manipulation (factorisation, rationalisation) resolve the cases where differentiation reproduces the form.
Can L'Hôpital's rule ever mislead about existence?
Yes: if f'/g' has no limit (oscillation), the rule is inconclusive — the original limit may still exist, as in lim (x + sin x)/x = 1 at infinity, settled by bounding.