L'Hospital's Rule

6 min read

The Rule That Turns Limits Into Derivatives

When direct substitution gives you $\frac{0}{0}$ or $\frac{\infty}{\infty}$, you’re stuck. The limit exists, but you can’t find it by plugging in values. L’Hospital’s Rule provides an elegant escape: differentiate the numerator and denominator separately, then try the limit again.

Why does this work? Intuitively, when both $f(x)$ and $g(x)$ approach 0 (or both approach $\infty$), the ratio $\frac{f(x)}{g(x)}$ depends on how fast each function approaches that value. Derivatives measure rates of change, exactly what we need.

This single technique handles most of the “hard” limits you’ll encounter in calculus.

Prerequisite Map

Quick Reference

Key Concepts

The Rule

L’Hospital’s Rule: Suppose $f$ and $g$ are differentiable and $g'(x) \neq 0$ near $a$ (except possibly at $a$). If

$$\lim_{x \to a} f(x) = 0 \text{ and } \lim_{x \to a} g(x) = 0$$

or

$$\lim_{x \to a} f(x) = \pm\infty \text{ and } \lim_{x \to a} g(x) = \pm\infty$$

then

$$\boxed{\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}}$$

provided the limit on the right exists (or equals $\pm\infty$).

Critical Points to Remember

1. Check the form FIRST. L’Hospital’s Rule only applies to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

2. Differentiate numerator and denominator SEPARATELY. This is NOT the quotient rule!

3. The rule extends to one-sided limits and limits at $\pm\infty$.

4. You may need to apply the rule multiple times if the result is still indeterminate.

5. After differentiating, always check if the new limit is determinate; do not automatically differentiate again.

Procedure

Step 1: Verify the limit has form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Step 2: Differentiate numerator and denominator separately: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Step 3: Evaluate the new limit:

• If determinate → you have the answer
• If still $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → apply L’Hospital’s Rule again
• If some other indeterminate form → try algebraic simplification

Common Mistake: Using the Quotient Rule

WRONG: $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x \cdot x - \sin x \cdot 1}{x^2}$ ✗

RIGHT: $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$ ✓

L’Hospital’s Rule says: differentiate top and bottom separately, not using the quotient rule.

Common Mistake: Applying When It Doesn’t Apply

WRONG: $$\lim_{x \to \pi} \frac{\sin x}{1 - \cos x} \stackrel{?}{=} \lim_{x \to \pi} \frac{\cos x}{\sin x}$$

This is wrong because at $x = \pi$:

• Numerator: $\sin \pi = 0$
• Denominator: $1 - \cos \pi = 1 - (-1) = 2 \neq 0$

The form is $\frac{0}{2}$, which is NOT indeterminate! The answer is simply $0$.

Always check the form before applying L’Hospital’s Rule.

Why It Works (Intuition)

Near $x = a$, if $f(a) = g(a) = 0$, both functions look approximately linear:

• $f(x) \approx f'(a)(x - a)$
• $g(x) \approx g'(a)(x - a)$

Their ratio is approximately: $$\frac{f(x)}{g(x)} \approx \frac{f'(a)(x-a)}{g'(a)(x-a)} = \frac{f'(a)}{g'(a)}$$

The $(x - a)$ terms cancel, leaving the ratio of derivatives.

        f(x)                     f'(a)·(x-a)
        ────  ≈  ─────────────  =  ─────────
        g(x)     g'(a)·(x-a)       g'(a)

         tangent line approximation

Worked Examples

Example 1: Basic $\frac{0}{0}$

Find $\lim_{x \to 1} \frac{\ln x}{x - 1}$.

Check form: As $x \to 1$: $\ln 1 = 0$ and $1 - 1 = 0$. Form: $\frac{0}{0}$ ✓

Apply L’Hospital’s Rule: $$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{1/x}{1} = \lim_{x \to 1} \frac{1}{x} = 1$$

Example 2: $\frac{\infty}{\infty}$ with Repeated Application

Find $\lim_{x \to \infty} \frac{e^x}{x^2}$.

Check form: As $x \to \infty$: $e^x \to \infty$ and $x^2 \to \infty$. Form: $\frac{\infty}{\infty}$ ✓

First application: $$\lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{e^x}{2x}$$

Check form again: Still $\frac{\infty}{\infty}$ ✓

Second application: $$\lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{e^x}{2} = \infty$$

Conclusion: $\lim_{x \to \infty} \frac{e^x}{x^2} = \infty$

This shows that $e^x$ grows faster than any polynomial!

Example 3: Simplify Instead of Re-applying

Find $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$.

Check form: $\frac{\infty}{\infty}$ ✓

First application: $$\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})}$$

Simplify instead of differentiating again: $$\frac{1/x}{1/(2\sqrt{x})} = \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}}$$

$$\lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0$$

Key insight: After one application, the expression simplified to something easy. Don’t blindly re-apply!

Practice Problems

Conceptual Questions (CCI-Style)

Common Misconceptions

Mastery Checklist

• I can verify that a limit has form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L’Hospital’s Rule
• I differentiate numerator and denominator separately (not using the quotient rule)
• I can apply L’Hospital’s Rule multiple times when needed
• I know when to simplify algebraically instead of applying the rule again
• I can recognize when L’Hospital’s Rule doesn’t apply
• I can explain why L’Hospital’s Rule works intuitively

Mental Model

The Speed Camera Analogy:

Imagine two cars (numerator and denominator) both approaching a finish line (0 or $\infty$). If they both cross at the same instant, who’s “winning”? You need a speed camera (derivative) to measure who’s approaching faster.

L’Hospital’s Rule replaces the question “who reached the finish line?” with “who was going faster?”, which actually determines the race’s outcome.

Connections

Looking back:

• Recognizing Indeterminate Forms tells us when to use this technique
• Derivative rules from MATH 161 are essential for applying the rule

Looking ahead:

• Converting Indeterminate Products shows how to convert other forms to use L’Hospital’s Rule
• Indeterminate Powers handles 0^0, infinity^0, 1^infinity forms

Last updated: 2026-01-22