L'Hôpital's Rule

The rule, stated precisely

Suppose f and g are differentiable on an open interval around a (except possibly at a itself), and g′(x) ≠ 0 there. If both

lim_x→a f(x) = lim_x→a g(x) = 0    or    lim_x→a |f(x)| = lim_x→a |g(x)| = ∞,

and lim f′(x) / g′(x) exists (or is ±∞), then

lim_x→a f(x) / g(x) = lim_x→a f′(x) / g′(x).

The rule applies just as well when a = ±∞ or when the limit is one-sided. The proof uses the Cauchy mean value theorem: there is a point c between x and a with [f(x) − f(a)]/[g(x) − g(a)] = f′(c)/g′(c), and as x → a the squeezed c drags the ratio with it.

Worked example: lim_x→0 sin(x)/x = 1

This is the most-quoted limit in calculus. Direct substitution gives 0/0 — indeterminate. Differentiate top and bottom:

lim_x→0 sin(x)/x = lim_x→0 cos(x)/1 = cos(0)/1 = 1.

The new limit is no longer indeterminate — direct substitution works. Done. (Pedants will point out that knowing d/dx sin x = cos x already presupposes this limit, so this is a circular proof of the limit itself. True — but L'Hôpital is still a valid verification, and the small-angle approximation sin x ≈ x for small x is the underlying geometric fact.)

Worked example: lim_x→∞ x²/eˣ = 0

Substitution gives ∞/∞. Differentiate twice:

lim x²/e^x = lim 2x/e^x = lim 2/e^x = 0.

This generalises: any polynomial divided by e^x has limit zero at infinity. Exponentials beat polynomials.

The seven indeterminate forms

Form	Example	How to handle
0/0	limx→0 sin(x)/x	Direct L'Hôpital — differentiate top and bottom.
∞/∞	limx→∞ ln(x)/x	Direct L'Hôpital.
0·∞	limx→0⁺ x·ln(x)	Rewrite as ln(x)/(1/x) → ∞/∞, then L'Hôpital.
∞−∞	limx→0 [1/sin(x) − 1/x]	Combine into one fraction → 0/0, then L'Hôpital.
0⁰	limx→0⁺ xˣ	Take log: ln L = lim x·ln(x) → use 0·∞ trick.
1^∞	limx→∞ (1+1/x)ˣ	Take log: ln L = lim x·ln(1+1/x) → 0·∞.
∞⁰	limx→∞ x^(1/x)	Take log: ln L = lim ln(x)/x → ∞/∞.

The last three are exponential forms. The trick is always the same: write y = f(x)^g(x), take ln on both sides to get ln y = g(x) ln f(x), evaluate that limit (which becomes 0·∞), then exponentiate to recover y.

Where the rule shows up

Physics limits. Small-amplitude approximations rely on lim_θ→0 sin(θ)/θ = 1 (pendulum, optics) and lim_v/c→0 γ = 1 (Newtonian limit of special relativity). Both are L'Hôpital-style cancellations of leading-order behaviour.

Asymptotic comparisons. Computer scientists ask whether n log n grows faster than n^1.5. The ratio (n log n)/n^1.5 = log n / n^0.5 tends to zero by L'Hôpital — so polynomials beat polylogs.

Defining (1 + 1/n)ⁿ → e. The 1^∞ form. Taking log gives n · ln(1 + 1/n); rewrite as ln(1 + 1/n) / (1/n); apply L'Hôpital to get 1; exponentiate to get e. This is the modern textbook construction of e.

Verifying derivatives at boundary points. Whether a piecewise-defined function has a derivative at a join point reduces to a 0/0 limit — exactly L'Hôpital's territory.

L'Hôpital vs related techniques

	L'Hôpital's rule	Algebraic manipulation	Squeeze theorem	Taylor expansion
Best when	0/0 or ∞/∞ with simple derivatives	Factor cancels cleanly	Bounded oscillation	Need leading-order behaviour
Worst when	Derivatives more complex than originals	No factoring possible	No tight bounds available	Function not analytic
Reusability	Apply repeatedly	Single shot	Single shot	Higher orders give more accuracy
Catches non-existent limits	No	Yes (gives concrete form)	Yes (no squeeze possible)	Yes (oscillating remainders)
Requires differentiability	Yes	No	No	Yes (smoothness)
Useful for sequences	After turning n into x	Yes	Yes	Indirectly

Common mistakes

• Applying L'Hôpital to a non-indeterminate form. If lim f/g already substitutes to 3/2 or 0/5, do not differentiate — you get the wrong answer. Direct substitution wins.
• Differentiating the quotient. The rule asks for f′ and g′ separately, not (f/g)′. The latter is the quotient rule and a different beast.
• Stopping too early. If f′/g′ is still 0/0, apply L'Hôpital again. Some rational limits need three or four passes.
• Stopping too late. Once you reach a determinate form, stop. Differentiating further gives spurious answers.
• Ignoring the side condition. The rule requires g′ ≠ 0 in a punctured neighbourhood. Functions with horizontal-tangent zeros violate this.
• Treating the rule as one-way. If f′/g′ has no limit, the original ratio may still have one. The classic counter-example is x → ∞ of (x + sin x)/x.