What is L'Hôpital's Rule? — BUders Math Concepts

Simple Definition

Some limits look like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ when you plug in the value directly — these are called indeterminate forms, and they're genuinely ambiguous. The ratio could approach any value, or diverge entirely. Direct substitution tells you nothing.

L'Hôpital's Rule (pronounced lo-pee-TAHL) resolves these by replacing the limit with a new one: take the derivative of the numerator and the derivative of the denominator separately (not the quotient rule!), then evaluate the limit of that new fraction. If it still gives an indeterminate form, apply the rule again.

Formal Statement

L'Hôpital's Rule

$$\text{If } \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\pm\infty}{\pm\infty}, \text{ then}$$ $$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

provided the right-hand limit exists (or is ±∞)

Indeterminate Forms

L'Hôpital's Rule directly applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$. Other forms need algebraic manipulation first:

0/0

Direct LH

∞/∞

Direct LH

0·∞

Rewrite as 0/0

∞−∞

Find common denom

0⁰

Take ln first

1^∞

Take ln first

Intuition

🏁

A race between two functions going to zero

Think of $\frac{f(x)}{g(x)}$ as a race between $f$ and $g$, both heading toward zero. Which one reaches zero faster? The limit tells you the ratio of their speeds. The derivative captures exactly that speed. So $\frac{f'}{g'}$ is the ratio of their instantaneous rates — who's winning the race to zero.

⚠️

Check the form first — every time

L'Hôpital's Rule only works on $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Applying it to a limit that gives $\frac{2}{5}$ (perfectly fine, just evaluate it!) will give a wrong answer. Always verify the indeterminate form before using the rule.

Step by Step

Evaluate $\lim_{x\to 0}\dfrac{\sin x}{x}$:

1

Check the form

At $x=0$: $\dfrac{\sin 0}{0} = \dfrac{0}{0}$ ✓ — indeterminate, LH applies.

2

Differentiate numerator and denominator separately

$\dfrac{d}{dx}[\sin x] = \cos x$ · $\dfrac{d}{dx}[x] = 1$

3

Evaluate the new limit

$\lim_{x\to 0}\dfrac{\cos x}{1} = \cos 0 = 1$

4

Conclude

$\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ ✓ — the famous limit, confirmed.

Quick Examples

Example 1 — 0/0

$\lim_{x\to 0}\dfrac{e^x - 1}{x}$

Form: 0/0 ✓
LH: $\dfrac{e^x}{1}$
At $x=0$: $e^0 = 1$

Example 2 — ∞/∞

$\lim_{x\to\infty}\dfrac{\ln x}{x}$

Form: ∞/∞ ✓
LH: $\dfrac{1/x}{1} = \dfrac{1}{x}$
$\to 0$ as $x\to\infty$

Example 3 — Apply twice

$\lim_{x\to 0}\dfrac{1-\cos x}{x^2}$

LH once: $\dfrac{\sin x}{2x}$ → still 0/0
LH again: $\dfrac{\cos x}{2}$
At 0: $\frac{1}{2}$

Example 4 — 0·∞ form

$\lim_{x\to 0^+} x\ln x$

Rewrite as $\dfrac{\ln x}{1/x}$ → ∞/∞
LH: $\dfrac{1/x}{-1/x^2} = -x$
$\to 0$

Common Mistakes

✗ Applying LH without an indeterminate form

If the limit gives $\frac{3}{5}$ directly, just evaluate it — don't apply L'Hôpital's Rule. Using LH on a non-indeterminate form gives an entirely wrong answer. Always check the form first.

✗ Using the quotient rule instead of separate derivatives

In L'Hôpital's Rule, you differentiate the numerator and denominator separately. Do NOT apply the quotient rule $\frac{f'g - fg'}{g^2}$. Just compute $f'$ and $g'$ independently and form the new fraction $\frac{f'}{g'}$.

✗ Applying LH once when the form is still indeterminate

If after one application you still get $\frac{0}{0}$, apply L'Hôpital's Rule again. You can repeat it as many times as needed, as long as each step produces an indeterminate form. Check the form after every application.

Related Concepts

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L'Hôpital's Rule appears in limits and indeterminate forms throughout Calculus I.

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