Difficulty: Intermediate | Prerequisites: Limits (Chapter 2), basic differentiation rules (Chapter 3)
This is the first of two sets of notes on Section 4.4. Part 1 covers the core rule and the two basic indeterminate forms (0/0 and ∞/∞), including repeated applications and cases where the rule fails. Part 2 covers indeterminate products, differences, and powers.
When direct substitution gives you 0/0 or ∞/∞, L'Hôpital's Rule lets you take the derivative of the top and the derivative of the bottom separately, then try the limit again. You can apply it more than once if the result is still indeterminate. The rule requires both functions to be differentiable and the denominator's derivative to be nonzero on the relevant interval.
Indeterminate form
An expression whose limit cannot be determined from the form alone, because the combination of limiting values could produce different results depending on the specific functions involved. Think of it as a "tie" between competing tendencies, where you need more information to break the tie.
Indeterminate form of type 0/0
A limit of the form lim f(x)/g(x) where both f(x) → 0 and g(x) → 0. In simple terms, both the numerator and denominator are shrinking to zero, and the final value depends on which one shrinks faster.
Indeterminate form of type ∞/∞
A limit of the form lim f(x)/g(x) where both f(x) → ±∞ and g(x) → ±∞. Both top and bottom are blowing up, and you need to work out which one grows faster.
L'Hôpital's Rule (also written L'Hospital's Rule)
If lim f(x)/g(x) produces a 0/0 or ∞/∞ indeterminate form, and if f and g are differentiable with g'(x) ≠ 0 on an open interval containing a (except possibly at a itself), then lim f(x)/g(x) = lim f'(x)/g'(x). In simple terms, you replace the original limit with the limit of the ratio of derivatives.
Differentiable
A function is differentiable at a point if its derivative exists there. For L'Hôpital's Rule, both f and g must be differentiable on an open interval around the point of interest. Think of it as the function being smooth enough to have a well-defined tangent line.
Open interval
An interval (a, b) that does not include its endpoints. L'Hôpital's Rule requires the functions to be differentiable on an open interval I containing the limit point, though the functions need not be differentiable at the limit point itself.
Back in Chapter 2, you learnt standard techniques for evaluating limits: direct substitution, factoring, rationalising. Those methods handle most limits, but they break down when you hit expressions like lim(x→0) (e^(2x) − 1)/x or lim(x→∞) (ln x − 1)/x. Direct substitution gives 0/0 or ∞/∞, and algebraic manipulation does not simplify things. L'Hôpital's Rule is the tool built for exactly these situations.
Suppose f and g are differentiable and g'(x) ≠ 0 on an open interval I that contains a (except possibly at a itself). Suppose that either:
lim(x→a) f(x) = 0 and lim(x→a) g(x) = 0 (the 0/0 case), or
lim(x→a) f(x) = ±∞ and lim(x→a) g(x) = ±∞ (the ∞/∞ case)
Then: lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x), provided the right-hand side exists (or is ±∞).
The rule also works for one-sided limits and for limits at ±∞. That is, "x → a" can be replaced by x → a⁺, x → a⁻, x → ∞, or x → −∞.
Before applying L'Hôpital's Rule, verify:
The limit is genuinely indeterminate (0/0 or ∞/∞). If substitution gives you something like 5/0 or 3/7, the rule does not apply.
Both f and g are differentiable near the point.
g'(x) ≠ 0 near the point (except possibly at the point itself).
Imagine two curves f(x) and g(x) that both pass through zero at x = a. If you zoom in close to the point (a, 0), the curves look approximately linear, like their tangent lines: f(x) ≈ f'(a)(x − a) and g(x) ≈ g'(a)(x − a). The ratio becomes f'(a)(x − a) / g'(a)(x − a) = f'(a)/g'(a), which is just the ratio of the derivatives. That is the heart of why the rule works.
When f(a) = g(a) = 0, f' and g' are continuous, and g'(a) ≠ 0, the proof is short. Using the alternative definition of the derivative:
lim(x→a) f'(x)/g'(x) = f'(a)/g'(a) = [lim(x→a) (f(x) − f(a))/(x − a)] / [lim(x→a) (g(x) − g(a))/(x − a)] = lim(x→a) (f(x) − f(a))/(g(x) − g(a)) = lim(x→a) f(x)/g(x)
The last step uses f(a) = g(a) = 0. The general proof (without the continuity assumption on the derivatives) is harder and typically covered in a real analysis course.
lim(x→0) (e^(2x) − 1)/x: Substitution gives 0/0. Differentiate top and bottom: lim(x→0) 2e^(2x)/1 = 2.
lim(x→∞) (ln x − 1)/x: Substitution gives ∞/∞. Differentiate: lim(x→∞) (1/x)/1 = 0.
lim(x→0) (sin x)/x: The classic 0/0 form. Differentiate: lim(x→0) cos x / 1 = 1. (You already knew this from Chapter 2, but L'Hôpital's Rule confirms it quickly.)
lim(x→1) (3x³ − 2x² + 4x − 5)/(4x⁴ − 2x − 2): Substitution gives 0/0. Differentiate: lim(x→1) (9x² − 4x + 4)/(16x³ − 2) = 9/14.
If, after one application of L'Hôpital's Rule, you still get 0/0 or ∞/∞, apply the rule again. Each time, verify the indeterminate form first.
lim(x→∞) (5eˣ − 1)/(3x²): Form ∞/∞. First application: lim(x→∞) 5eˣ/(6x). Still ∞/∞. Second application: lim(x→∞) 5eˣ/6 = ∞. The exponential grows faster than any polynomial.
lim(x→0) (eˣ − x − 1)/(3x²): Form 0/0. First application: lim(x→0) (eˣ − 1)/(6x). Still 0/0. Second application: lim(x→0) eˣ/6 = 1/6.
lim(x→0) (tan x − x)/x³: Form 0/0. First: lim(x→0) (sec²x − 1)/(3x²). Still 0/0. Second: lim(x→0) (2 sec²x tan x)/(6x). Still 0/0. Third: evaluate carefully to get 1/3.
The limit is not indeterminate. Example: lim(x→∞) (e^(2x) − 1)/x gives ∞/∞, so the rule applies. But lim(x→π⁻) (sin x)/(1 − cos x) gives 0/2, which is just 0. No rule needed, and applying it anyway would give a wrong answer.
The derivative ratio has no limit. Example: lim(x→0) x sin(1/x). This is a 0·∞ product, not a quotient. Even if you rearrange it into a quotient, the derivative ratio may oscillate without settling on a value. The Squeeze Theorem works here instead.
The rule goes in circles. Example: lim(x→∞) √(x² + 1)/x. Differentiating gives lim(x→∞) x/√(x² + 1), which is the reciprocal of the original. Applying the rule again brings you back to the start. Instead, divide top and bottom by x to get √(1 + 1/x²), which approaches 1.
You can evaluate directly. Example: lim(x→0) eˣ/(eˣ + e⁻ˣ). Substitution gives 1/2 immediately. Applying L'Hôpital's Rule here is unnecessary and technically invalid (there is no indeterminate form).
L'Hôpital's Rule
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}Valid when the left side is an indeterminate form of type 0/0 or ∞/∞, and f, g are differentiable with g'(x) ≠ 0 near a.
Special-case proof chain (when f(a) = g(a) = 0)
\lim_{x \to a} \frac{f'(x)}{g'(x)} = \frac{f'(a)}{g'(a)} = \frac{\displaystyle\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\displaystyle\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \lim_{x \to a} \frac{f(x)}{g(x)}Indeterminate forms come up in engineering whenever you need the limiting behaviour of a ratio. In electrical engineering, the sinc function sin(x)/x (which equals 1 at x = 0 by L'Hôpital's Rule) is fundamental to signal processing and filter design. In physics, L'Hôpital's Rule appears when deriving the behaviour of systems at boundary conditions, for instance working out what happens to a damped oscillator's response as the damping ratio approaches a critical value.
Students often apply L'Hôpital's Rule when the limit is not indeterminate. If substitution gives you a determinate value like 5/3 or 0/7, the rule does not apply, and using it will produce a wrong answer.
Students sometimes use the quotient rule (for differentiating f/g) instead of differentiating the numerator and denominator separately. L'Hôpital's Rule says to take f'(x)/g'(x), not [f/g]'(x). These are completely different.
Students forget to re-check the indeterminate form before each repeated application. Just because the first application was valid does not mean the second one is. Always verify 0/0 or ∞/∞ again.
Students assume the rule always works. If differentiating leads to a ratio that oscillates or cycles back to the original, the rule has failed for that problem, and a different technique (algebraic simplification, Squeeze Theorem, direct evaluation) is needed.
Expect at least one problem requiring multiple applications of L'Hôpital's Rule (e.g. exponential vs polynomial growth).
A classic exam trap: a limit that looks like it needs L'Hôpital's Rule but can be evaluated directly, or one that is not actually indeterminate. Read the problem before reaching for derivatives.
The limit lim(x→0) (sin x)/x = 1 appears constantly as a building block in other problems. Know it cold.
Problems may ask you to identify when L'Hôpital's Rule cannot be applied. The circular-result case (√(x² + 1)/x) is a favourite.
Historical note for context: the rule was first published by L'Hôpital in 1696, but was actually discovered by Johann Bernoulli in 1694. L'Hôpital purchased the rights to Bernoulli's discoveries. This sometimes appears as a bonus or trivia question.
True or False: L'Hôpital's Rule can be applied to any limit involving a fraction.
Answer: False. It only applies when the limit produces an indeterminate form of type 0/0 or ∞/∞.
Fill in the blank: When applying L'Hôpital's Rule, you differentiate the numerator and denominator ______, not using the quotient rule.
Answer: separately.
True or False: If one application of L'Hôpital's Rule still gives 0/0, you may apply it again.
Answer: True, provided the conditions (differentiability, g'(x) ≠ 0) still hold.
Fill in the blank: L'Hôpital's Rule is valid for one-sided limits and for limits at ______.
Answer: infinity (±∞).
True or False: lim(x→0) eˣ/(eˣ + e⁻ˣ) requires L'Hôpital's Rule.
Answer: False. Substitution gives 1/2 directly.
Q: Evaluate lim(x→0) (e^(2x) − 1)/x.
A: Form 0/0. Apply L'Hôpital's Rule: lim(x→0) 2e^(2x)/1 = 2(1) = 2.
Q: Evaluate lim(x→∞) (ln x − 1)/x.
A: Form ∞/∞. Apply L'Hôpital's Rule: lim(x→∞) (1/x)/1 = 0. Logarithmic growth is slower than linear.
Q: Evaluate lim(x→∞) (5eˣ − 1)/(3x²).
A: Form ∞/∞. First application: lim(x→∞) 5eˣ/(6x), still ∞/∞. Second application: lim(x→∞) 5eˣ/6 = ∞. Exponential growth dominates polynomial growth.
Q: Evaluate lim(x→0) (eˣ − x − 1)/(3x²).
A: Form 0/0. First application: lim(x→0) (eˣ − 1)/(6x), still 0/0. Second application: lim(x→0) eˣ/6 = 1/6.
Q: Can you apply L'Hôpital's Rule to lim(x→π⁻) (sin x)/(1 − cos x)? Explain.
A: As x → π⁻, sin x → 0 and 1 − cos x → 1 − (−1) = 2. This gives 0/2 = 0, which is not indeterminate. L'Hôpital's Rule does not apply. The answer is simply 0.
Q: Evaluate lim(x→0) (tan x − x)/x³.
A: Form 0/0. First: lim(x→0) (sec²x − 1)/(3x²), still 0/0. Second: lim(x→0) (2 sec²x tan x)/(6x), still 0/0. Factor out: (1/3) lim(x→0) (sec²x)(tan x / x). Since tan x / x → 1 and sec²(0) = 1, the answer is 1/3.
Q: Why does L'Hôpital's Rule fail for lim(x→∞) √(x² + 1)/x?
A: Differentiating gives lim(x→∞) (x/√(x² + 1))/1 = lim(x→∞) x/√(x² + 1), which is the reciprocal of the original limit. Applying the rule again cycles back. Instead, divide top and bottom by x: √(1 + 1/x²) → 1.
L'Hôpital's Rule sits at the intersection of limits (Chapter 2) and derivatives (Chapter 3). The geometric intuition, that the ratio of functions near a point equals the ratio of their tangent-line slopes, is really a statement about linear approximation, which connects to Taylor series later in the course.
The rule also connects forward to improper integrals (Chapter 7), where you will need to evaluate limits of the form ∞/∞ when determining convergence. In differential equations, indeterminate forms arise when analysing equilibrium solutions and stability.
Part 2 of these notes covers the remaining indeterminate forms: products (0 · ∞), differences (∞ − ∞), and powers (0⁰, ∞⁰, 1^∞), all of which are converted into 0/0 or ∞/∞ so that L'Hôpital's Rule can be applied.
L'Hôpital's Rule, L'Hospital's Rule, L'Hôpital's Theorem, lopital rule, indeterminate forms, 0/0 form, infinity over infinity, ∞/∞, zero over zero, limit of a quotient, ratio of derivatives, repeated L'Hôpital, when L'Hôpital fails, Calculus II limits, Section 4.4, NYU Calculus II for Engineers, Johann Bernoulli, indeterminate limit evaluation, derivative limit rule