- L’Hôpital’s rule says that if a limit gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the top and bottom separately and try the limit again.
- It is a calculus shortcut for limits that algebra (factorising, rationalising) cannot easily reach, such as \( \frac{\sin x}{x} \) or \( \frac{e^x – 1 – x}{x^2} \).
- If the new limit is still \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you may apply the rule again, checking the form each time.
- In our case, a 24-month plan of $500 payments has a present value that tends to exactly $12,000 as the interest rate tends to zero.
- Exam point: always show that the limit is an indeterminate form before using the rule. Using it on a limit that is not \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) gives a wrong answer.
The case: what is a payment plan worth when interest is almost zero?
A furniture shop offers a sofa on a payment plan: 24 monthly payments of $500. A buyer wants to compare this with paying cash. To do that, she works out the present value of the plan, the amount of money today that is worth the same as the 24 future payments. With a monthly interest rate \( r \) (as a decimal), the present value of an ordinary annuity is
She tries a few rates in a spreadsheet. At 2% a month the plan is worth about $9,457 today. At 1% it is worth about $10,622. At 0.5% it is about $11,281. As the rate falls, the present value climbs. Common sense says that at zero interest the plan is worth exactly what you pay, \( 24 \times 500 = 12000 \) dollars.
But when she types \( r = 0 \) into the formula, the spreadsheet shows an error. The top is \( 1 – 1^{-24} = 0 \) and the bottom is 0. The formula gives \( \frac{0}{0} \), which is not a number. So is the answer really $12,000, and can we prove it? This is exactly the kind of question L’Hôpital’s rule answers. (The figures are illustrative.)
The concept: limits of the form 0/0 and ∞/∞
A limit asks what value a function gets close to as the input gets close to some number. Usually you just substitute. But sometimes substitution gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are called indeterminate forms. “Indeterminate” means the form alone does not tell you the answer. The limit could be 0, 5, 12,000, infinity, or it might not exist at all. It depends on how fast the top and the bottom approach their values.
That last idea is the heart of the rule. Near the point, if both top and bottom are heading to zero, their ratio is decided by their rates of change. If the top is shrinking twice as fast as the bottom, the ratio settles near 2. Rates of change are derivatives. So we compare the derivatives instead of the functions.
A quick picture helps. Near \( x = a \), a smooth function \( f \) with \( f(a) = 0 \) looks like its tangent line, \( f(x) \approx f'(a)(x – a) \). Do the same for \( g \). Then
The \( (x – a) \) factors cancel, which is exactly what factorising does in the algebra method. L’Hôpital’s rule turns that idea into a theorem [1].
Here \( f(x) \) is the numerator and \( g(x) \) is the denominator. \( f'(x) \) and \( g'(x) \) are their derivatives, taken separately. This is not the quotient rule: you do not differentiate the fraction as a whole. The number \( a \) can be a finite value, or \( +\infty \) or \( -\infty \). The condition “if the new limit exists” matters: the rule only lets you move from the derivative limit back to the original limit, not the other way round.
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The rule is named after the French mathematician Guillaume de l’Hôpital, who published it in 1696 in the first textbook on differential calculus. The result itself is usually credited to his teacher, Johann Bernoulli [1]. You will also see it spelled L’Hospital, the older French spelling.
Why this method fits the case
Our present-value limit is a textbook \( \frac{0}{0} \). Can algebra do it? You could expand \( (1 + r)^{-24} \) with the binomial series, but that needs an infinite series and is easy to get wrong. You cannot simply factorise \( r \) out of the top. L’Hôpital’s rule needs just one derivative of the top and one of the bottom. It is the most direct route.
More generally, L’Hôpital’s rule fits when the top and bottom mix different kinds of function: exponentials with polynomials, logs with powers, trig functions with polynomials. These are hard to cancel by algebra. It also handles growth-rate comparisons at infinity, such as “does \( x^{2} \) or \( e^{0.5x} \) grow faster?”, which matter in economics when comparing polynomial costs with exponential growth.
When is it not the right tool?
- When the limit is not indeterminate. \( \lim_{x \to 0} \frac{x + 1}{x + 2} = \frac{1}{2} \) by substitution. Using L’Hôpital would wrongly give \( \frac{1}{1} = 1 \).
- When simple algebra is quicker. \( \frac{x^2 – 9}{x – 3} \) factorises to \( x + 3 \), giving 6. Our existing post on evaluating 0/0 limits with algebra covers factorising and rationalising, which are often faster and are what some syllabuses expect.
- When the derivatives get worse, not better. For \( \frac{\sqrt{x^2 + 1}}{x} \) as \( x \to \infty \), each application flips the fraction and you go round in circles. Dividing top and bottom by \( x \) gives the answer, 1, immediately.
- When the derivative limit does not exist. More on this below: the rule is then silent, not wrong, and you need another method.
Setting it up
Name the top and bottom, check the form, then differentiate each one separately.
| Piece | Expression | Value at \( r = 0 \) | Derivative with respect to \( r \) | Derivative at \( r = 0 \) |
|---|---|---|---|---|
| Numerator \( f(r) \) | \( 500\left[1 – (1 + r)^{-24}\right] \) | 0 | \( 500 \times 24(1 + r)^{-25} = 12000(1 + r)^{-25} \) | 12,000 |
| Denominator \( g(r) \) | \( r \) | 0 | 1 | 1 |
| Form | \( \frac{0}{0} \): indeterminate, rule applies | New limit \( \frac{12000}{1} = 12000 \) | ||
The derivative of \( -(1 + r)^{-24} \) uses the chain rule: bring down the \(-24\), reduce the power by one, and multiply by the derivative of \( 1 + r \), which is 1. The two minus signs make a plus.
Solving it
Show that \( \lim_{r \to 0} 500 \cdot \frac{1 – (1 + r)^{-24}}{r} = 12000 \).
As the monthly rate tends to zero, the plan’s present value tends to exactly $12,000, the plain sum of the payments.
Repeated use. Find \( \lim_{x \to 0} \frac{e^x – 1 – x}{x^2} \).
The limit is \( \frac12 \), which matches the \( x^2 \) coefficient in the Maclaurin series \( e^x = 1 + x + \frac{x^2}{2} + \dots \).
An \( \frac{\infty}{\infty} \) limit. A cost grows like \( x^2 \) while a fund grows like \( e^{0.5x} \). Find \( \lim_{x \to \infty} \frac{x^2}{e^{0.5x}} \).
The limit is 0: any exponential with a positive rate eventually outgrows any power of \( x \).
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What the answer tells us
The result confirms common sense: with no interest, money today and money later are worth the same, so 24 payments of $500 are worth $12,000. But the limit tells us more than the value at one point. It tells us the present-value formula has a removable gap at \( r = 0 \). If we define \( PV(0) = 12000 \), the curve becomes continuous. Figure 1 shows the curve heading smoothly into that point. Spreadsheet models of loans often add an “if rate = 0” line for exactly this reason.
Is the answer reasonable? Yes. Present value must be less than $12,000 for any positive rate, because each payment is discounted. The table of values, $9,457 at 2%, $10,622 at 1%, $11,281 at 0.5% and $11,985 at 0.01%, rises steadily towards 12,000 and never passes it.
How sensitive is the value to the rate near zero? We can apply the same ideas to find the slope at \( r = 0 \). A careful expansion gives \( PV(r) \approx 12000 – 150000r \) for small \( r \). At \( r = 0.001 \) (0.1% a month) that predicts $11,850, and the exact value is $11,851.29. So each 0.1 percentage point of monthly interest knocks about $150 off the value of the plan. For the buyer, that is the real lesson. A “0% finance” deal is worth noticeably more than one at even a small rate, and the gap grows with the number of payments.
Finally, the rule can fail to give an answer. Take \( \lim_{x \to \infty} \frac{x + \sin x}{x} \). It is \( \frac{\infty}{\infty} \). L’Hôpital gives \( \lim_{x \to \infty} (1 + \cos x) \), which swings between 0 and 2 and has no limit. But the original limit does exist: divide by \( x \) to get \( 1 + \frac{\sin x}{x} \to 1 \). The rule says “if the derivative limit exists, the original equals it”. When the derivative limit does not exist, the rule tells you nothing, and you must use another method [1].
For more on limits, continuity and differentiation in one place, the Mathematics Made Simple textbook builds the ideas step by step with graphs. The Mathematics Made Simple Practice Questions book gives you plenty of limit questions with full worked answers.
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Practice questions
For each limit, first check the form, then decide whether L’Hôpital’s rule is needed.
EasyQ1. Find \( \lim_{x \to 2} \frac{x^2 – 4}{x – 2} \) using L’Hôpital’s rule, then check by factorising. Show worked answer
L’Hôpital: \( \lim_{x \to 2} \frac{2x}{1} = 4 \).
Check: \( \frac{(x – 2)(x + 2)}{x – 2} = x + 2 \to 4 \). ✓
EasyQ2. Find \( \lim_{x \to 0} \frac{\sin 3x}{x} \). Show worked answer
L’Hôpital: \( \lim_{x \to 0} \frac{3\cos 3x}{1} = 3 \).
MediumQ3. Find \( \lim_{x \to 0} \frac{1 – \cos x}{x^2} \). Show worked answer
Once: \( \lim_{x \to 0} \frac{\sin x}{2x} \), still \( \frac{0}{0} \).
Twice: \( \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2} \).
MediumQ4. Find \( \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} \). Show worked answer
L’Hôpital: \( \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0 \).
So \( \sqrt{x} \) grows faster than \( \ln x \).
ExamQ5. Find \( \lim_{x \to 0} \frac{e^{2x} – 1 – 2x}{x \sin x} \). Justify each use of L’Hôpital’s rule. [5] Show worked answer
Once: \( \frac{2e^{2x} – 2}{\sin x + x\cos x} \). [1] Top \( \to 0 \), bottom \( \to 0 \): still \( \frac{0}{0} \). [1]
Twice: \( \frac{4e^{2x}}{2\cos x – x\sin x} \). [1]
At \( x = 0 \): \( \frac{4}{2} = 2 \). [1]
Frequently asked questions
Can I use L’Hôpital’s rule for 0 × ∞ or ∞ − ∞?
Not directly. First rewrite the expression as a fraction. For example, \( x\ln x \) as \( x \to 0^+ \) is \( 0 \times (-\infty) \); write it as \( \frac{\ln x}{1/x} \), which is \( \frac{-\infty}{\infty} \), and then apply the rule to get 0.
What about limits like \( (1 + \tfrac{1}{n})^n \)?
That is the form \( 1^\infty \). Take logs to get \( n\ln(1 + \tfrac{1}{n}) \), rewrite it as \( \frac{\ln(1 + 1/n)}{1/n} \), which is \( \frac{0}{0} \), and use the rule. The log tends to 1, so the original limit is \( e \).
Is L’Hôpital’s rule the same as the algebra method for 0/0?
They answer the same question in different ways. Algebra removes the common factor that makes both parts zero. L’Hôpital compares the rates at which the parts reach zero. For polynomials and roots, algebra is often quicker; for exponentials, logs and trig, L’Hôpital usually is.
Can I use L’Hôpital’s rule to prove \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)?
In an exam answer it gives the right number. But as a proof it is circular, because the derivative of \( \sin x \) is itself found using that very limit. Textbooks prove it with a geometric squeeze argument instead.
References
- Strang, G. and Herman, E. (2016) Calculus Volume 1. Houston, TX: OpenStax. Available at: https://openstax.org/details/books/calculus-volume-1.
- College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description. New York: College Board.
- International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021). Cardiff: IBO.
- Sydsæter, K., Hammond, P., Strøm, A. and Carvajal, A. (2016) Essential Mathematics for Economic Analysis. 5th edn. Harlow: Pearson.
