Limits · Topic 07 of 08

L'Hôpital's Rule — The Shortcut for Indeterminate Forms

When a limit produces 0/0 or ∞/∞, algebra often fails. L'Hôpital's Rule provides a powerful shortcut: differentiate the numerator and denominator separately, then try again. Used correctly, it resolves limits that would otherwise take pages of manipulation.

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§ 01The Statement of L'Hôpital's Rule

Named after the French mathematician Guillaume de l'Hôpital (1661–1704), this rule transforms a ratio of functions into a ratio of their derivatives: but only under specific conditions.

Suppose f and g are differentiable on an open interval containing a (except possibly at a itself), and that g′(x) ≠ 0 near a. If the limit of f(x)/g(x) as x → a produces an indeterminate form (specifically 0/0 or ∞/∞) then:

L'Hôpital's Rule
limx→a f(x)/g(x) = limx→a f′(x)/g′(x)
provided the right-hand limit exists (or equals ±∞),
and the original limit is of the form 0/0 or ∞/∞

This looks deceptively simple: but every word in the conditions matters. Let's unpack exactly what is required before we can apply it.

Historical Note Despite the name, the rule was actually discovered by Johann Bernoulli, who communicated it to L'Hôpital under a paid tutoring arrangement. L'Hôpital published it in 1696 in the first ever calculus textbook. The attribution has stuck ever since.

§ 02The Four Conditions — When You CAN Use It

L'Hôpital's Rule is not a universal limit shortcut. It applies only when all four of the following conditions hold:

Checklist Before Applying L'Hôpital's Rule

  1. The form is 0/0 or ∞/∞. These are the only two forms the rule directly handles. Other indeterminate forms (0·∞, ∞−∞, 1∞, 0⁰, ∞⁰) must first be converted into 0/0 or ∞/∞.
  2. Both f and g are differentiable near a. If either function has a corner, cusp, or is not differentiable near a, the rule may not apply.
  3. g′(x) ≠ 0 near a (except possibly at a). If g′ = 0 throughout a neighbourhood of a, the rule fails.
  4. The new limit lim f′/g′ must exist (or equal ±∞). If the differentiated limit also diverges in an oscillatory way, the rule is inconclusive, not wrong, just unhelpful.
The #1 Mistake — Applying the Rule When the Form is NOT Indeterminate L'Hôpital's Rule only works on indeterminate forms. If substituting x = a gives something like 3/5 or 7/0, do NOT differentiate. The rule will give a wrong answer. Always verify the form first.
Wrong ApplicationMisapplying the rule to limx→2 (x²+1)/(x+3)

Substituting x = 2 gives 5/5 = 1. This is a determinate form: direct substitution works perfectly.

❌ Incorrect: Using L'Hôpital gives 2x/1 → 4/1 = 4. This is wrong.
✓ Correct: Direct substitution gives (4+1)/(2+3) = 5/5 = 1

§ 03Applying the Rule — 0/0 Forms

The most common application is resolving 0/0 indeterminate forms that resist factoring or other algebraic approaches. The three-step process is: confirm the form, differentiate top and bottom separately, evaluate.

3-Step Process

  1. Confirm: Substitute x = a and verify you get 0/0 or ∞/∞.
  2. Differentiate: Find f′(x) and g′(x) separately. Do NOT use the quotient rule, this is not d/dx[f/g].
  3. Evaluate: Take limx→a f′(x)/g′(x). If still indeterminate, apply the rule again.
Example 1limx→0 sin(x)/x

Substituting x = 0: sin(0)/0 = 0/0. ✓ Indeterminate, apply the rule.

  1. f(x) = sin(x) → f′(x) = cos(x)
  2. g(x) = x → g′(x) = 1
  3. limx→0 cos(x)/1 = cos(0)/1 = 1
limx→0 sin(x)/x = 1
Example 2limx→0 (eˣ − 1)/x

At x = 0: (1 − 1)/0 = 0/0. ✓

  1. f′(x) = eˣ, g′(x) = 1
  2. limx→0 eˣ/1 = e⁰ = 1
limx→0 (eˣ − 1)/x = 1
Example 3limx→0 (1 − cos x)/x²

At x = 0: 0/0. ✓

  1. f′(x) = sin x, g′(x) = 2x → still 0/0 at x = 0. Apply again.
  2. f′′(x) = cos x, g′′(x) = 2
  3. limx→0 cos x / 2 = 1/2
limx→0 (1 − cos x)/x² = 1/2
Example 4limx→1 (ln x)/(x − 1)

At x = 1: ln(1)/(1−1) = 0/0. ✓

  1. f′(x) = 1/x, g′(x) = 1
  2. limx→1 (1/x)/1 = 1/1 = 1
limx→1 ln(x)/(x − 1) = 1
Example 5limx→0 (x − sin x)/x³

At x = 0: 0/0. Apply repeatedly until resolved.

  1. 1st application: (1 − cos x)/(3x²) → 0/0 at x = 0.
  2. 2nd application: sin x/(6x) → 0/0 at x = 0.
  3. 3rd application: cos x/6 → cos(0)/6 = 1/6
limx→0 (x − sin x)/x³ = 1/6

§ 04The ∞/∞ Form — Comparing Rates of Growth

L'Hôpital's Rule applies equally to ∞/∞ forms, making it the fastest way to establish the dominance hierarchy: which functions grow faster as x → ∞.

Example 6limx→∞ ln(x)/x

As x → ∞: ln x → ∞ and x → ∞. Form is ∞/∞. ✓

  1. f′(x) = 1/x, g′(x) = 1
  2. limx→∞ (1/x)/1 = limx→∞ 1/x = 0
limx→∞ ln(x)/x = 0 — polynomial dominates logarithm
Example 7limx→∞ x²/eˣ

Both → ∞. Apply the rule twice.

  1. 1st: 2x/eˣ, still ∞/∞.
  2. 2nd: 2/eˣ → 2/∞ = 0
limx→∞ x²/eˣ = 0 — exponential dominates any polynomial
Example 8limx→∞ (3x² + 5x)/(2x² − x + 1)

Both polynomials → ∞. Apply the rule twice (or just compare leading terms).

  1. 1st: (6x + 5)/(4x − 1): still ∞/∞.
  2. 2nd: 6/4 = 3/2
limx→∞ (3x²+5x)/(2x²−x+1) = 3/2
Dominance Hierarchy (slowest to fastest growth)
ln x ≪ x^p ≪ eˣ ≪ x! ≪ xˣ

L'Hôpital confirms: any ratio of adjacent types → 0 (slower on top) or ∞ (faster on top).

§ 05Converting Other Indeterminate Forms

The other five indeterminate forms cannot be fed directly into L'Hôpital's Rule: they must first be rearranged into 0/0 or ∞/∞. Here is the conversion strategy for each.

0/0
Direct ✓
∞/∞
Direct ✓
0 · ∞
Rewrite as fraction
∞ − ∞
Common denominator
1∞
ln trick
0⁰
ln trick
∞⁰
ln trick

Converting 0 · ∞

Rewrite the product as a fraction by moving one factor to the denominator as its reciprocal. Choose the reciprocal that creates a 0/0 or ∞/∞ form.

0 · ∞ Conversion

  1. f(x) · g(x) where f → 0 and g → ∞
  2. Rewrite as: f(x) / (1/g(x)) → 0/0   or   g(x) / (1/f(x)) → ∞/∞
  3. Apply L'Hôpital to the resulting fraction
Example 9limx→0⁺ x · ln(x)

Form: 0 · (−∞). Rewrite x as 1/(1/x):

  1. limx→0⁺ ln(x) / (1/x), form is −∞/∞. ✓
  2. f′(x) = 1/x, g′(x) = −1/x²
  3. (1/x) / (−1/x²) = (1/x) · (−x²) = −x
  4. limx→0⁺ (−x) = 0
limx→0⁺ x · ln(x) = 0

Converting 1∞, 0⁰, and ∞⁰ — The ln Trick

When the limit has the form [f(x)]g(x) and produces 1∞, 0⁰, or ∞⁰, take the natural logarithm first. Since ln is continuous, the limit of ln[fg] = limit of g·ln(f), which is a 0·∞ form we can then convert to a fraction.

The ln Trick for Exponential Indeterminate Forms

  1. Let L = lim [f(x)]g(x)
  2. Take ln: ln L = lim g(x) · ln f(x), now a 0 · ∞ form
  3. Convert to a fraction and apply L'Hôpital
  4. Get the value of ln L, call it k
  5. Therefore L = ek
Example 10limx→∞ (1 + 1/x)ˣ — the definition of e

Form: 1∞. Apply the ln trick.

  1. Let y = (1 + 1/x)ˣ. Then ln y = x · ln(1 + 1/x), form 0·∞ (since 1/x → 0, x → ∞).
  2. Rewrite: ln(1 + 1/x) / (1/x), form 0/0. ✓
  3. Differentiate top: d/dx ln(1+1/x) = (−1/x²)/(1+1/x) = −1/(x²+x)
  4. Differentiate bottom: d/dx (1/x) = −1/x²
  5. Ratio: [−1/(x²+x)] / [−1/x²] = x²/(x²+x) = x/(x+1)
  6. limx→∞ x/(x+1) = 1, so ln L = 1, therefore L = e1 = e
limx→∞ (1 + 1/x)ˣ = e
Example 11limx→0⁺ xˣ — the 0⁰ form

Form: 0⁰. Apply the ln trick.

  1. Let y = xˣ. Then ln y = x · ln x, this is 0·(−∞), handled in Example 9.
  2. From Example 9: limx→0⁺ x · ln x = 0
  3. So ln L = 0, therefore L = e⁰ = 1
limx→0⁺ xˣ = 1

Converting ∞ − ∞

Combine into a single fraction over a common denominator. This usually converts the form to 0/0.

Example 12limx→0 (1/sin x − 1/x)

Form: ∞ − ∞. Combine over common denominator x·sin x.

  1. (x − sin x) / (x · sin x), form 0/0 at x = 0. ✓
  2. Apply L'Hôpital: (1 − cos x) / (sin x + x cos x), still 0/0.
  3. Apply again: sin x / (cos x + cos x − x sin x) = sin x / (2cos x − x sin x)
  4. Substitute x = 0: 0 / (2 − 0) = 0
limx→0 (1/sin x − 1/x) = 0

§ 06Applying the Rule Repeatedly

If after one application of L'Hôpital's Rule the new limit is still an indeterminate form, you may apply the rule again. There is no limit to how many times you can repeat it: as long as each new form is 0/0 or ∞/∞ and the differentiability conditions remain satisfied.

Example 13limx→0 (eˣ − 1 − x − x²/2) / x³

At x = 0: 0/0. Apply up to three times.

  1. 1st: (eˣ − 1 − x) / (3x²) → 0/0
  2. 2nd: (eˣ − 1) / (6x) → 0/0
  3. 3rd: eˣ / 6 → e⁰/6 = 1/6
limx→0 (eˣ − 1 − x − x²/2)/x³ = 1/6
When Repeated Application Fails — Circular Limits Sometimes L'Hôpital creates a cycle: lim f/g leads to lim g/f leads back to lim f/g. This happens with limits like limx→∞ eˣ/eˣ (which is trivially 1) and some combinations of exponentials. When you notice cycling, stop and use a different approach — factoring, substitution, or recognising the pattern directly.

§ 07Further Worked Examples

Example 14limx→π sin(x)/(π − x)

At x = π: sin(π)/(π−π) = 0/0. ✓

  1. f′(x) = cos(x), g′(x) = −1
  2. limx→π cos(x)/(−1) = cos(π)/(−1) = (−1)/(−1) = 1
limx→π sin(x)/(π − x) = 1
Example 15limx→0⁺ xsin x — the ∞⁰ / 0⁰ form

As x → 0⁺: x → 0 and sin x → 0. Form: 0⁰. Apply the ln trick.

  1. ln y = sin(x) · ln(x), form 0 · (−∞). Rewrite as ln(x)/csc(x) → −∞/∞. ✓
  2. f′(x) = 1/x, g′(x) = −csc(x)cot(x)
  3. Ratio: (1/x) / (−csc x cot x) = −sin(x)tan(x)/x
  4. As x → 0⁺: sin(x)/x → 1 and tan(x) → 0, so the product → 1 · 0 = 0
  5. ln L = 0, therefore L = e⁰ = 1
limx→0⁺ xsin x = 1

§ 08Common Mistakes

MistakeWhat Goes WrongFix
Using the rule on a determinate form lim (x²+1)/(x+3) at x=2 is 5/5 = 1, not indeterminate. Applying the rule gives the wrong answer 4. Always substitute first. Only proceed with L'Hôpital if you get 0/0 or ∞/∞.
Applying the Quotient Rule instead of differentiating separately d/dx[f/g] ≠ f′/g′. The quotient rule gives (f′g − fg′)/g². L'Hôpital's Rule uses f′/g′ — the derivatives of the numerator and denominator independently. Differentiate numerator alone, denominator alone. Never cross-multiply.
Forgetting to check the conditions after each application After one application, the new form might not be 0/0 or ∞/∞ anymore — direct substitution may now work. After each application, substitute and check. Stop applying as soon as the form is determinate.
Applying the rule directly to 0·∞, 1∞, or ∞−∞ These forms are not covered by the rule. Applying it directly gives nonsense. Convert to 0/0 or ∞/∞ first using the algebraic techniques in §05.
Concluding a limit doesn't exist when lim f′/g′ oscillates If lim f′/g′ doesn't exist (e.g. oscillates), L'Hôpital is inconclusive — the original limit may still exist. Try a completely different method. L'Hôpital's Rule failing is not evidence the limit fails.

§ 09Quick Reference — Key Results via L'Hôpital

LimitFormResult
limx→0 sin(x)/x0/01
limx→0 (eˣ−1)/x0/01
limx→0 (1−cos x)/x²0/0 twice1/2
limx→1 ln(x)/(x−1)0/01
limx→0 (x−sin x)/x³0/0 three times1/6
limx→∞ ln(x)/x∞/∞0
limx→∞ xⁿ/eˣ (any n)∞/∞ n times0
limx→0⁺ x·ln(x)0·∞ → 0/00
limx→∞ (1+1/x)ˣ1∞ → ln tricke
limx→0⁺ xˣ0⁰ → ln trick1

§ 10Practice Quiz — L'Hôpital's Rule

10 questions spanning the rule statement, conditions, 0/0, ∞/∞, and converted forms. Immediate feedback on every answer.

Score: 0 / 0
Q1 of 10

L'Hôpital's Rule states lim f/g = lim f′/g′. Which forms does it directly apply to?

L'Hôpital's Rule directly handles only 0/0 and ∞/∞. All other indeterminate forms (0·∞, ∞−∞, 1∞, 0⁰, ∞⁰) must first be converted into one of these two forms.

Q2 of 10

Evaluate: limx→0 sin(3x) / x

Form 0/0. f′ = 3cos(3x), g′ = 1. At x=0: 3cos(0)/1 = 3·1 = 3.

Q3 of 10

A student evaluates limx→3 (x²+1)/(x−1) using L'Hôpital and gets 2x/1 = 6. What is wrong?

Substituting x=3: (9+1)/(3−1) = 10/2 = 5. This is determinate. L'Hôpital cannot be used here and gives the wrong answer. The correct answer is 5.

Q4 of 10

Evaluate: limx→0 (eˣ − 1 − x) / x²

  1. Form 0/0. 1st application: (eˣ−1)/(2x), still 0/0.
  2. 2nd application: eˣ/2. At x=0: 1/2 = 0.5.
Q5 of 10

Which is the correct first step to evaluate limx→0⁺ x · ln(x)?

0·∞ cannot be fed directly into L'Hôpital. Rewrite as ln(x)/(1/x) to get −∞/∞, which the rule handles. Differentiating: (1/x)/(−1/x²) = −x → 0.

Q6 of 10

Evaluate: limx→∞ ln(x) / √x

Form ∞/∞. f′ = 1/x, g′ = 1/(2√x). Ratio: (1/x)/(1/(2√x)) = 2√x/x = 2/√x → 0 as x→∞.

Q7 of 10

To evaluate limx→∞ (1 + 2/x)x using the ln trick, we get ln L = lim x·ln(1+2/x). What does this limit equal?

Rewrite as ln(1+2/x)/(1/x) → 0/0. Differentiate: top = (−2/x²)/(1+2/x), bottom = −1/x². Ratio → 2 as x→∞. So ln L = 2, meaning L = e².

Q8 of 10

Evaluate: limx→0 (tan x − x) / x³

  1. 0/0. 1st: (sec²x − 1)/(3x²) = tan²x/(3x²) → 0/0
  2. 2nd: 2tan x · sec²x / (6x) → 0/0
  3. 3rd: (2sec⁴x + 4tan²x·sec²x)/6. At x=0: 2(1)+0)/6 = 2/6 = 1/3 ≈ 0.333
Q9 of 10

If after applying L'Hôpital's Rule the new limit lim f′/g′ oscillates and does not exist, what can you conclude about lim f/g?

L'Hôpital is inconclusive when lim f′/g′ fails to exist. The original limit may still exist: a classic example is limx→∞ (x + sin x)/x = 1, even though after differentiation lim (1 + cos x)/1 oscillates.

Q10 of 10

What is limx→0⁺ xˣ?

Form 0⁰. Let y = xˣ. ln y = x·ln x → 0 (from Example 9). So ln L = 0, giving L = e⁰ = 1.

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Quiz complete!

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