Pure mathematics

L'Hôpital's Rule: a rule for determining limits

L'Hôpital's rule is usually filed under an optional Further Pure module and consequently ignored by most students. This is a shame: the limits it resolves appear throughout A-level Maths and compulsory Further Pure, and the rule is often the cleanest — sometimes the only — way to evaluate them.

L'Hôpital's rule

If substituting x = a into f(x)/g(x) gives the indeterminate form 0/0 or ∞/∞, then — provided f and g are differentiable near a and the limit on the right exists:

Differentiate the numerator and denominator separately — this is not the quotient rule. Then evaluate the new limit. If the result is still indeterminate, apply the rule again.

Indeterminate forms

Not every expression of the form 0/0 or ∞/∞ is truly indeterminate — the point is that direct substitution fails to determine the limit and L'Hôpital's rule is needed. The table below lists the forms the rule handles directly, and those that require a preliminary rewriting step before the rule can be applied.

Form	Example	Strategy
0/0	sin(x)/x as x→0	Apply rule directly
∞/∞	ln(x)/x as x→∞	Apply rule directly
0 · ∞	x·ln(x) as x→0⁺	Rewrite as fraction first
∞ − ∞	1/x − 1/sin(x) as x→0	Combine over a common denominator

The 0/0 form — limits from A-level Maths

A-level MathsEdexcelAQAOCROCR MEI

The following limits arise directly in A-level Maths — they underpin the standard derivative results for sin x and e^x. Without L'Hôpital's rule, they require geometric arguments or squeeze theorems. With it, they are one line.

Example 1 — the limit behind differentiating sin x from first principles

Direct substitution gives 0/0. Apply L'Hôpital:

Try it yourself

Use L'Hôpital's rule to evaluate lim_x→0 tan(x) / x.

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Direct substitution gives 0/0. Apply L'Hôpital:

Example 2 — the limit behind differentiating eˣ from first principles

Try it yourself

Evaluate lim_x→0 (e^2x − 1) / x.

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Example 3 — a polynomial limit (A-level Maths, factorisation not needed)

Although (x² − 9)/(x − 3) can be factorised as (x + 3), L'Hôpital gives the answer without any factorisation:

Try it yourself

Evaluate lim_x→2 (x² − 4) / (x − 2) without factorising.

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Example 4 — the limit behind differentiating ln x from first principles

Try it yourself

Evaluate lim_x→0 ln(1 + 3x) / x.

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The ∞/∞ form — growth rate comparisons

A-level Further Maths — Further Pure (compulsory)Edexcel (Core Pure 2)AQA (Further Pure 2)OCR (Pure Core 2)OCR MEI (Core Pure B)

These limits establish the relative growth rates of the standard functions — a standard result in compulsory Further Pure.

Example 5 — logarithm grows slower than any power of x

Try it yourself

Evaluate lim_x→∞ ln(x) / x².

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Example 6 — exponential grows faster than any polynomial

Each application of the rule reduces the degree of the numerator by one. After three steps the numerator is constant and the limit is clearly zero. This is the systematic proof that e^x dominates every polynomial.

Try it yourself

Evaluate lim_x→∞ x² / e^x.

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Forcing a fraction — the rewriting trick for 0 · ∞

L'Hôpital's rule requires a fraction. If the expression is instead a product u(x)·v(x) tending to 0·∞, it is not immediately in the right form. The fix is simple: rewrite the product as a fraction by moving one factor into the denominator as its reciprocal.

Both rewritings are valid fractions; choose whichever produces the simpler derivative. The rule can then be applied.

Example 7 — x ln x as x → 0⁺

Direct substitution: 0 · (−∞). Place ln x over 1/x to create an ∞/∞ form:

Why place ln x on top rather than x? Placing x on top would give a 0/0 form with derivative 1/(ln x)⁻¹ — considerably messier. Choosing which factor to invert is part of the technique.

Try it yourself

Evaluate lim_x→0+ x² ln x.

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Direct substitution: 0 · (−∞). Write as ln(x) / x⁻²:

Example 8 — x e⁻ˣ as x → ∞

Direct substitution: ∞ · 0. Place e^x in the denominator:

Try it yourself

Evaluate lim_x→∞ x² e^−x.

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Example 9 — 1/x − 1/sin x as x → 0 (∞ − ∞ form)

Direct substitution: ∞ − ∞. Rather than inverting a factor, combine over a common denominator to create a single fraction:

Now both numerator (→ 0) and denominator (→ 0) give a 0/0 form:

Try it yourself

Evaluate lim_x→0 (1/x − 1/tan x). Combine over a common denominator first.

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Both numerator and denominator → 0. Apply L'Hôpital:

Applying the rule recursively

After one application of L'Hôpital's rule, the result may still be indeterminate. Provided the conditions hold for f'/g', the rule can be applied again — and again — until a determinate limit is reached.

Example 10 — (1 − cos x)/x² — two applications

The first application gives sin x / 2x, which is still 0/0. Only after the second application does a definite value emerge.

Try it yourself

Evaluate lim_x→0 (1 − cos 3x) / x².

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Example 11 — (eˣ − 1 − x)/x² — two applications

Try it yourself

Evaluate lim_x→0 (e^2x − 1 − 2x) / x².

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Example 12 — (eˣ − 1 − x − x²/2)/x³ — three applications

Each differentiation peels off one layer of the expression until only e^xremains.

Try it yourself

How many applications of L'Hôpital's rule are needed to evaluate lim_x→0 (cos x − 1 + x²/2) / x⁴? Find the value.

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Four applications, peeling off one derivative each time:

Consistent with the Maclaurin series for cos x: cos x ≈ 1 − x²/2 + x⁴/24 − …, so the coefficient of x⁴ is 1/24.

What the pattern is telling you

Notice the denominators: 2, 6 = 3!, 24 = 4!, … These are exactly the denominators in the Maclaurin series for e^x. Repeated application of L'Hôpital's rule to the expression

is precisely computing the next Maclaurin coefficient f⁽ⁿ⁾(0)/n!. L'Hôpital's rule and the Maclaurin series are two faces of the same differentiation-at-zero idea.

A caution — when not to use the rule

L'Hôpital's rule requires both numerator and denominator to be differentiable in a punctured neighbourhood of a, and the limit of f'/g' must exist (or be ±∞). Two common mistakes:

Applying the rule when the form is not indeterminate. For example, lim_x→0 (x + 1)/x² → ∞ directly; L'Hôpital would give 1/(2x) → ∞ too, but there was never any need for it.
Confusing L'Hôpital with the quotient rule. Differentiate the numerator and denominator separately:

Where the rule appears in the specifications

A-level Further MathsEdexcel (Further Pure 2)AQA (Further Pure 2)OCR (Further Pure B)OCR MEI (Further Pure A)

L'Hôpital's rule is formally assessed only in the optional Further Pure units listed above. However, this does not mean it is irrelevant to everyone else. Consider the limits it resolves:

Limit	Where it arises	Normally proved by
sin(x)/x → 1	A-level Maths: differentiating sin x	Geometric / squeeze theorem
(e^x−1)/x → 1	A-level Maths: differentiating e^x	Definition of e, series argument
ln(1+x)/x → 1	A-level Maths: differentiating ln x	Definition of ln, series argument
ln(x)/x → 0 as x→∞	Further Core Pure: growth rate results	Comparison argument
xⁿ/e^x → 0 as x→∞	Further Core Pure: growth rate results	Induction / comparison

In each case, L'Hôpital gives the result in one or two lines where the standard proof requires a separate, longer argument. Even if it will not appear on your exam paper by name, knowing it makes many limits that appear throughout the course essentially trivial.