The rule
Before you start: you'll need standard derivatives (see the Derivatives guide). The rule itself is proved using Cauchy's mean value theorem.
Suppose that, as \(x \to a\),
1. An indeterminate form
\(f(x) \to 0\) and \(g(x) \to 0\), or \(f(x) \to \pm\infty\) and \(g(x) \to \pm\infty\);
2. Differentiability
\(f\) and \(g\) are differentiable near \(a\), with \(g'(x) \ne 0\) near \(a\) (except possibly at \(a\) itself);
3. The new limit exists
\(\displaystyle\lim_{x \to a}\frac{f'(x)}{g'(x)}\) exists (or is \(\pm\infty\)).
Conclusion
\[\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}\]
The rule also works when \(a\) is \(\infty\) or \(-\infty\), and for one-sided limits.
Important: differentiate the top and the bottom separately. This is not the quotient rule.
Why it works
Take the \(\tfrac{0}{0}\) case with \(f(a) = g(a) = 0\). By Cauchy's mean value theorem on \([a, x]\),
\[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)} \quad \text{for some } c \text{ between } a \text{ and } x\]
As \(x \to a\), \(c\) is squeezed towards \(a\) too, so \(\dfrac{f(x)}{g(x)}\) and \(\dfrac{f'(c)}{g'(c)}\) have the same limit.
Intuitively: near \(a\), both functions are almost straight lines through zero, \(f(x) \approx f'(a)(x - a)\) and \(g(x) \approx g'(a)(x - a)\). Their ratio is then just the ratio of their slopes.
The method
- Substitute first. If you get an ordinary number, that's the answer. Use the rule only for \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\).
- Differentiate the numerator and the denominator separately.
- Substitute again. If it's still \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\), repeat. Check the form every time.
- Simplify between steps where you can, using trig identities or standard limits such as \(\dfrac{\sin x}{x} \to 1\). Repeated differentiation can grow messy.
For the other indeterminate forms (\(0\cdot\infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), \(\infty^0\)), first rewrite them as \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\). That's the next lesson: indeterminate forms.
Solved examples
Example 1: The classic
EasyEvaluate \(\displaystyle\lim_{x \to 0}\frac{\sin x}{x}\).
- At \(x = 0\) the form is \(\tfrac{0}{0}\).
- Differentiate top and bottom: \(\dfrac{\cos x}{1}\).
- Substitute: \(\cos 0 = 1\).
Answer\(1\)
Example 2: Applying the rule twice
EasyEvaluate \(\displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}\).
- Form \(\tfrac{0}{0}\). Differentiate: \(\dfrac{e^x - 1}{2x}\), still \(\tfrac{0}{0}\).
- Differentiate again: \(\dfrac{e^x}{2}\).
- Substitute: \(\dfrac{1}{2}\).
Answer\(\dfrac{1}{2}\)
Example 3: Three times
MediumEvaluate \(\displaystyle\lim_{x \to 0}\frac{x - \sin x}{x^3}\).
- \(\tfrac{0}{0}\): differentiate to get \(\dfrac{1 - \cos x}{3x^2}\), still \(\tfrac{0}{0}\).
- Again: \(\dfrac{\sin x}{6x}\), still \(\tfrac{0}{0}\).
- Again: \(\dfrac{\cos x}{6} \to \dfrac{1}{6}\). (Or stop one step earlier and use \(\dfrac{\sin x}{x} \to 1\).)
Answer\(\dfrac{1}{6}\)
Example 4: An ∞/∞ limit
EasyEvaluate \(\displaystyle\lim_{x \to \infty}\frac{\log x}{x}\).
- Form \(\tfrac{\infty}{\infty}\). Differentiate: \(\dfrac{1/x}{1} = \dfrac{1}{x}\).
- As \(x \to \infty\), \(\dfrac{1}{x} \to 0\).
Answer\(0\). Logarithms grow more slowly than any power of \(x\).
Example 5: Exponentials beat powers
MediumEvaluate \(\displaystyle\lim_{x \to \infty}\frac{x^3}{e^x}\).
- \(\tfrac{\infty}{\infty}\). Differentiate three times: \(\dfrac{3x^2}{e^x}\), then \(\dfrac{6x}{e^x}\), then \(\dfrac{6}{e^x}\).
- \(\dfrac{6}{e^x} \to 0\).
Answer\(0\)
Example 6: Simplify along the way
Exam levelEvaluate \(\displaystyle\lim_{x \to 0}\frac{\tan x - x}{x - \sin x}\).
- \(\tfrac{0}{0}\). Differentiate: \(\dfrac{\sec^2 x - 1}{1 - \cos x} = \dfrac{\tan^2 x}{1 - \cos x}\).
- Instead of differentiating again, simplify: \(\dfrac{\tan^2 x}{1 - \cos x} = \dfrac{\sin^2 x}{\cos^2 x\,(1 - \cos x)} = \dfrac{(1 - \cos x)(1 + \cos x)}{\cos^2 x\,(1 - \cos x)} = \dfrac{1 + \cos x}{\cos^2 x}\).
- Substitute \(x = 0\): \(\dfrac{2}{1}\).
Answer\(2\)
Example 7: When the rule says nothing
Exam levelEvaluate \(\displaystyle\lim_{x \to \infty}\frac{x + \sin x}{x}\).
- Form \(\tfrac{\infty}{\infty}\). Differentiating gives \(\dfrac{1 + \cos x}{1}\), which keeps oscillating between 0 and 2. It has no limit, so condition 3 fails and the rule tells us nothing.
- Divide by \(x\) instead: \(1 + \dfrac{\sin x}{x}\). Since \(|\sin x| \le 1\), \(\dfrac{\sin x}{x} \to 0\).
Answer\(1\). The limit exists even though L'Hôpital's rule couldn't find it.
Common mistakes
1. Using the rule when the form isn't indeterminate
\(\displaystyle\lim_{x \to 1}\frac{x^2 + 1}{x + 1} = \frac{2}{2} = 1\) by direct substitution. Wrongly differentiating gives \(\dfrac{2x}{1} \to 2\). Always substitute first.
2. Using the quotient rule
L'Hôpital's rule differentiates top and bottom separately: \(\dfrac{f'}{g'}\), not \(\left(\dfrac{f}{g}\right)'\).
3. Not re-checking the form after each step
After one differentiation you might already have an ordinary number. Differentiating again then gives a wrong answer.
4. Concluding the limit doesn't exist when \(\dfrac{f'}{g'}\) has no limit
As Example 7 shows, the original limit can still exist. The rule simply doesn't apply.
How it's asked in exams
- Evaluate a limit of the form \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\), often needing two or three applications.
- Find constants \(a\), \(b\) so that a limit has a given value.
- Combined with other forms: \(0\cdot\infty\), \(\infty - \infty\) and exponential forms first rewritten into \(\tfrac{0}{0}\). See indeterminate forms.
Write the form, \(\left(\tfrac{0}{0}\right)\) or \(\left(\tfrac{\infty}{\infty}\right)\), next to every step. It shows the examiner you checked, and it stops you differentiating one step too many.
Practice questions
Try each one on paper before you open the answer.
Q1\(\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x^2}\)
\(\tfrac{0}{0}\) → \(\dfrac{\sin x}{2x} \to \dfrac{1}{2}\).
Q2\(\displaystyle\lim_{x \to 0}\frac{e^x - e^{-x}}{\sin x}\)
\(\tfrac{0}{0}\) → \(\dfrac{e^x + e^{-x}}{\cos x} \to \dfrac{2}{1} = 2\).
Q3\(\displaystyle\lim_{x \to 0}\frac{\log(1 + x)}{x}\)
\(\tfrac{0}{0}\) → \(\dfrac{1/(1 + x)}{1} \to 1\).
Q4\(\displaystyle\lim_{x \to \pi/2}\frac{1 - \sin x}{\cos x}\)
\(\tfrac{0}{0}\) → \(\dfrac{-\cos x}{-\sin x} \to \dfrac{0}{1} = 0\).
Q5\(\displaystyle\lim_{x \to 0}\frac{a^x - b^x}{x}\)
\(\tfrac{0}{0}\) → \(\dfrac{a^x\log a - b^x\log b}{1} \to \log a - \log b = \log\dfrac{a}{b}\).
Q6\(\displaystyle\lim_{x \to 0}\frac{\sin x - x\cos x}{x^3}\)
\(\tfrac{0}{0}\) → \(\dfrac{\cos x - \cos x + x\sin x}{3x^2} = \dfrac{\sin x}{3x} \to \dfrac{1}{3}\).
Q7\(\displaystyle\lim_{x \to \infty}\frac{x^2}{e^{2x}}\)
\(\tfrac{\infty}{\infty}\) → \(\dfrac{2x}{2e^{2x}}\) → \(\dfrac{2}{4e^{2x}} \to 0\).
Q8Challenge. \(\displaystyle\lim_{x \to 0}\frac{x e^x - \log(1 + x)}{x^2}\)
\(\tfrac{0}{0}\) → \(\dfrac{e^x + xe^x - \frac{1}{1 + x}}{2x}\), still \(\tfrac{0}{0}\).
Again → \(\dfrac{2e^x + xe^x + \frac{1}{(1 + x)^2}}{2} \to \dfrac{2 + 0 + 1}{2} = \dfrac{3}{2}\).
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Frequently asked questions
When can I use L'Hôpital's rule?
Only when direct substitution gives \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\). Other indeterminate forms must be rewritten into one of these first.
Is L'Hôpital's rule the same as the quotient rule?
No. You differentiate the numerator and denominator separately and divide. The quotient rule finds the derivative of a fraction, which is a different thing.
How many times can I apply it?
As many times as needed, provided the form is still \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\) each time.
Why is it sometimes spelled L'Hospital?
Both spellings refer to the same French mathematician, Guillaume de l'Hôpital. “L'Hospital” is the older spelling, and many Indian textbooks still use it.
Where L'Hôpital's rule leads next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.