Unit 4: Contextual Applications of Differentiation

L'Hôpital's Rule

Evaluating indeterminate-form limits using L'Hôpital's Rule.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Local Linearization

Using Derivatives to Resolve a Stubborn Limit

limₓ→0 sin(x)/x gives 0/0 on direct substitution, and no factoring or rationalizing simplifies it the way earlier 0/0 limits did. Before reading on: could differentiating the numerator and denominator separately — not as a quotient, just each piece on its own — reveal the limit some other way?

Definition — L'Hôpital's Rule

If limₓ→c f(x)/g(x) produces the indeterminate form 0/0 or ∞/∞, and f and g are differentiable near c, then limₓ→c f(x)/g(x) = limₓ→c f'(x)/g'(x), provided that second limit exists. The rule can be applied again if the new limit is still indeterminate.

\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)} \quad (\text{if } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty})

Worked Example — Resolving a Classic 0/0 Limit

Evaluate limₓ→0 sin(x)/x using L'Hôpital's Rule. Direct substitution gives sin(0)/0 = 0/0 — indeterminate. Differentiate numerator and denominator separately: d/dx[sin x] = cos x, d/dx[x] = 1. limₓ→0 cos(x)/1 = cos(0) = 1. This recovers the well-known fundamental trig limit, now via derivatives instead of a geometric argument.

Worked Example — Resolving an ∞/∞ Limit, Applying the Rule Twice

Evaluate limₓ→∞ (3x² + 5x)/(2x² − 1). Direct substitution gives ∞/∞. Differentiate: (6x + 5)/(4x). This is still ∞/∞ as x → ∞, so apply the rule again: 6/4 = 3/2. Check independently by dividing numerator and denominator by x²: (3 + 5/x)/(2 − 1/x²) → 3/2 as x → ∞ — matching.

Tip

Confirm the form is genuinely 0/0 or ∞/∞ by direct substitution before differentiating anything — L'Hôpital's Rule doesn't apply to any other indeterminate-looking situation, and applying it where it doesn't belong produces a confidently wrong answer.

Common Mistakes

Applying L'Hôpital's Rule to limₓ→0 (x + 1)/x, which gives 1/0 on direct substitution, not 0/0.
1/0 is not an indeterminate form the rule applies to — differentiating anyway gives 1/1 = 1, a confidently wrong answer, since the true limit doesn't even exist (the function goes to +∞ from the right and −∞ from the left). Always confirm 0/0 or ∞/∞ first.
Applying the quotient rule to f(x)/g(x) instead of differentiating the numerator and denominator separately.
L'Hôpital's Rule replaces the whole fraction with a new fraction of separate derivatives, f'(x)/g'(x) — it is not the quotient rule applied to f/g, and mixing the two produces an entirely different (and incorrect) expression.

Key Takeaways

L'Hôpital's Rule replaces an indeterminate 0/0 or ∞/∞ limit with the limit of the derivatives of the numerator and denominator, taken separately.
The rule can be applied repeatedly as long as each new limit is still indeterminate.
The rule only applies to genuine 0/0 or ∞/∞ forms — applying it elsewhere produces a wrong answer even though the differentiation itself is valid.

Summary

This closes Unit 4: related rates, linearization, and L'Hôpital's Rule all put the derivative to work in applied and limit-resolving contexts. Unit 5 turns back to a function's own graph, using the derivative to fully analyze its shape.

Track your progress

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