🎯 Limits & Continuity
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L'Hôpital's Rule

Resolve indeterminate forms using the ratio of derivatives.

L'Hôpital's Rule states: if lim f(x)/g(x) produces an indeterminate form 0/0 or ∞/∞, then lim f(x)/g(x) = lim f'(x)/g'(x), provided the latter limit exists.

If lim⁡x→af(x)g(x)=00 or ∞∞, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\text{If } \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\infty}{\infty}, \text{ then } \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}
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You differentiate the numerator and denominator separately — NOT the quotient rule. Apply L'Hôpital's Rule only when you have a true indeterminate form.

Indeterminate Forms Handled

0/0: direct form∞/∞: direct form0·∞: rewrite as 0/(1/∞) or ∞/(1/0)1^∞, 0⁰, ∞⁰: use logarithms first∞ − ∞: combine into a fraction first
✏️Apply L'Hôpital's to sin(x)/x as x → 0
Direct substitution gives 0/0. Apply L'Hôpital: lim sin(x)/x = lim cos(x)/1 = cos(0)/1 = 1. This confirms the famous limit lim_{x→0} sin(x)/x = 1.
lim⁡x→0sin⁡xx=L′Hlim⁡x→0cos⁡x1=1\lim_{x\to 0}\frac{\sin x}{x} \overset{L'H}{=} \lim_{x\to 0}\frac{\cos x}{1} = 1
✏️Repeated Application
lim_{x→0} (eˣ − 1 − x)/x². Both num and denom → 0. Apply once: (eˣ − 1)/(2x) → 0/0 again. Apply again: eˣ/2 → e⁰/2 = 1/2.
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Remember This!

L'Hôpital's Rule can be applied repeatedly. Keep applying until the limit is no longer indeterminate. But check that each application is valid (each step must still give 0/0 or ∞/∞).

✏️ Try It!

Evaluate lim_{x→∞} x/eˣ using L'Hôpital's Rule.

Practice with CalcVerse
→Limit Simulator
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