L'Hôpital's Rule
Evaluates indeterminate limits 0/0 or ∞/∞.
Formula
Variables
Example
lim(x→0) sin x / x = lim cos x / 1 = 1.
How it works
L'Hôpital's Rule lets you replace a difficult quotient of two functions with a quotient of their derivatives when the original limit gives an indeterminate form like 0/0 or ∞/∞. The idea comes from comparing how fast the numerator and denominator change near the limit point; if both vanish or both blow up, their slopes can reveal the ratio of their local growth rates. It only applies when the functions are differentiable near the point and the derivative quotient has a limit, so it is a tool for evaluation, not a universal shortcut.
When to use it
Use it when a limit of a fraction turns into 0/0 or ∞/∞ after direct substitution, especially in advanced calculus problems involving exponentials, logarithms, trig functions, or combinations of these. It is also useful in asymptotic comparisons, series-based limit problems, and rate-of-growth questions in applied math and physics.
Worked example, step by step
Evaluate lim(x→0) (e^x - 1) / x Step 1: Substitute x = 0: (e^0 - 1)/0 = 0/0, so L'Hôpital's Rule applies. Step 2: Differentiate numerator and denominator separately: d/dx(e^x - 1) = e^x, and d/dx(x) = 1. Step 3: Take the new limit: lim(x→0) e^x / 1. Step 4: Substitute x = 0 again: e^0 / 1 = 1. Final answer: lim(x→0) (e^x - 1) / x = 1.
Common mistakes to avoid
- Using L'Hôpital's Rule when the limit is not an indeterminate form like 0/0 or ∞/∞
- Differentiating the quotient with the quotient rule instead of differentiating numerator and denominator separately
- Stopping after one application even though the new quotient is still indeterminate
Did You Know?
L'Hôpital's rule was actually discovered by Johann Bernoulli — L'Hôpital just paid to publish it.
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