L'Hôpital's Rule (AA HL)
When a limit gives the indeterminate form \(\tfrac00\) or \(\tfrac{\infty}{\infty}\), l'Hôpital's rule lets you differentiate the numerator and denominator separately and try the limit again. It's fast once you spot the form, but it's also easy to misapply. This page covers the rule, when it does and doesn't apply, and the mistakes that lose marks. It's part of the broader Series & Limits topic.
23 questions on this sub-topic.
The rule
Covered under IB syllabus reference AHL5.13: evaluating limits of the form \(\lim_{x\to a}\tfrac{f(x)}{g(x)}\) using l'Hôpital's rule or a Maclaurin series, for the indeterminate forms \(\tfrac00\) and \(\tfrac{\infty}{\infty}\), including repeated use where the first application still gives an indeterminate form.
L'Hôpital's rule
\(\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}\)
Valid only when direct substitution gives \(\tfrac00\) or \(\tfrac{\infty}{\infty}\). Not in the formula booklet.
Want the full topic overview and GDC guidance for this area? See Series & Limits.
Worked examples
Evaluate \(\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x}\).
Worked solution
Form \(\dfrac{0}{0}\): apply L'Hôpital. M1
\(\displaystyle\lim_{x \to 0} \frac{3\cos 3x}{1}\) M1
\(= 3\cos 0 = 3\) A1
Evaluate \(\displaystyle\lim_{x \to 0^+} x\ln x\).
Worked solution
Rewrite as \(\dfrac{\ln x}{1/x}\), which is \(\dfrac{-\infty}{\infty}\) as \(x \to 0^+\). M1
Apply L'Hôpital: \(\displaystyle\lim_{x \to 0^+} \frac{1/x}{-1/x^2}\) M1
\(= \displaystyle\lim_{x \to 0^+} \frac{x^2/x}{-1} = \lim_{x \to 0^+}(-x)\) M1
\(\displaystyle\lim_{x \to 0^+} x \ln x = 0\) A1
Evaluate the following limits using l'H\(\hat{\text{o}}\)pital's rule.
(a) \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x}\)
(b) \(\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}\)
Worked solution
(a) Form \(\tfrac{0}{0}\), apply l'H\(\hat{\text{o}}\)pital: \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}x = 1\) M1
\(\displaystyle\lim_{x\to 0}\frac{\cos x}{1} = 1\) A1
(b) Form \(\tfrac{0}{0}\): \(\displaystyle\lim_{x\to 0}\frac{e^x}{1} = \frac{e^0}{1}\) M1
\( = 1\) A1
Evaluate \(\displaystyle\lim_{x \to 0^+} x^x\).
Worked solution
Let \(L = \displaystyle\lim_{x \to 0^+} x^x\). Take natural logs: \(\ln L = \displaystyle\lim_{x \to 0^+} x \ln x\). M1
Rewrite: \(\displaystyle\lim_{x \to 0^+} \frac{\ln x}{1/x}\) - form \(\dfrac{-\infty}{\infty}\). Apply L'Hôpital. M1
\(\displaystyle\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+}(-x)\) M1
\(= 0\) A1
So \(\ln L = 0\)
\(L = e^0 = 1\), so \(\displaystyle\lim_{x \to 0^+} x^x = 1\). A1
Common mistakes
- Applying l'Hôpital's rule without checking the form first. The rule only applies to \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) - differentiating top and bottom of a limit that isn't indeterminate gives a meaningless answer, even though the steps look identical.
- Stopping after one application when the form is still indeterminate. If differentiating once still leaves \(\tfrac00\) or \(\tfrac{\infty}{\infty}\), differentiate again - the rule can be reapplied as many times as needed, checking the form each time.
- Differentiating the whole fraction with the quotient rule instead. L'Hôpital's rule differentiates the numerator and denominator separately and forms a new fraction - it is not the same operation as the quotient rule for differentiating \(f(x)/g(x)\) itself.
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Quick answers
When can you use l'Hôpital's rule?
Only when direct substitution gives an indeterminate form of \(\tfrac00\) or \(\tfrac{\infty}{\infty}\). You must check and state the form before differentiating - the rule does not apply otherwise.
What do you do if l'Hôpital's rule still gives an indeterminate form?
Differentiate the numerator and denominator again and re-check the form. The rule can be applied repeatedly as long as each new limit is still \(\tfrac00\) or \(\tfrac{\infty}{\infty}\).