L’Hôpital’s Rule
In short: L’Hôpital’s rule evaluates a limit that looks like 0/0 or infinity/infinity by replacing the top and bottom with their derivatives and taking the limit of that new fraction.
AP Calculus ABCalculusAP Calculus AB Unit 4
What is L'Hôpital's Rule?
L’Hôpital’s rule rescues limits where direct substitution turns a fraction into \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\): the answer is hidden, not missing. It compares how fast the top and the bottom are heading toward their targets.
If \(f\) and \(g\) are differentiable near \(c\) and the quotient has one of those two indeterminate forms at \(x = c\), then
\[\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f^{\prime}(x)}{g^{\prime}(x)}\]
provided the right-hand limit exists. Note that this is the ratio of derivatives, not the derivative of the ratio.
How to spot it
- Plugging in the limit value gives 0/0 or infinity/infinity, and factoring or rationalizing does not help.
- The limit mixes a trig function with a polynomial, such as (1 – cos 4x) divided by x squared.
- A limit at infinity compares two growing expressions and you must decide which one wins.
- The wording asks you to “evaluate the limit” and a quick substitution fails.
The method
- Substitute the limit value first and confirm the form is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\); if not, the rule is not allowed.
- Differentiate the numerator and the denominator separately, with no quotient rule.
- Evaluate the new limit; if it is still indeterminate, apply L’Hôpital’s rule again.
- State the final value and mention the rule so the grader sees the justification.
L'Hôpital's Rule: solved example
Problem
Evaluate \(\lim_{x \to 0}\frac{1 – \cos 4x}{x^2}\).
Solution
At \(x = 0\) the form is \(\frac{1 – 1}{0} = \frac{0}{0}\), so the rule applies.
Differentiate top and bottom: \(\frac{4\sin 4x}{2x}\). This is still \(\frac{0}{0}\) at \(x = 0\), so apply it again: \(\frac{16\cos 4x}{2}\).
Now substitution works: \(\frac{16 \cdot 1}{2} = \) 8.
Try it
Question 1
Evaluate \(\lim_{x \to 2}\frac{x^3 – 8}{x^2 – 4}\).
Show the answer
It is \(\frac{0}{0}\). Differentiating gives \(\frac{3x^2}{2x}\), which at \(x = 2\) equals \(\frac{12}{4} = 3\).
Question 2
Can the rule be used on \(\lim_{x \to 1}\frac{x + 3}{x – 1}\)?
Show the answer
No. Substitution gives \(\frac{4}{0}\), which is not an indeterminate form; the limit does not exist as a finite number (the values blow up).
L'Hôpital's Rule vs. Quotient Rule
The quotient rule finds the derivative of \(\frac{f}{g}\) and produces a squared denominator. L’Hôpital’s rule leaves the fraction alone and differentiates the top and bottom separately, then takes a limit. Mixing the two is the most common slip on these problems.
Keep going
Lessons
Worksheets and quizzes
- Linearization and L’Hopital’s Rule – Worksheet
- Linearization and L’Hopital’s Rule – Quiz
Related words
- Derivative
- Limit Definition of the Derivative
- Quotient Rule
- Intermediate Value Theorem
- Derivative Rules
Whole dictionary: A–Z · AP Calculus AB · Calculus