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.
| Property | Value |
|---|---|
| Section | Stewart 6.8 |
| Course | MATH162 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
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 $\pm\infty$).
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:
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.
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 $x = \pi$:
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.
Near $x = a$, if $f(a) = g(a) = 0$, both functions look approximately linear:
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 approximationFind $\lim_{x \to 1} \frac{\ln x}{x - 1}$.
Check form: As $x \to 1$: $\ln 1 = 0$ and $1 - 1 = 0$. Form: $\frac{0}{0}$ ✓
Apply L'Hospital's Rule: $$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{1/x}{1} = \lim_{x \to 1} \frac{1}{x} = 1$$
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!
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!
Evaluate each limit using L'Hospital's Rule:
Evaluate $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.
Evaluate each limit:
What happens if you try to use L'Hospital's Rule on $\displaystyle\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}}$?
Find the limit using a different method.
Prove that for any positive integer $n$:
$$\lim_{x \to \infty} \frac{e^x}{x^n} = \infty$$
What does this tell us about the relative growth rates of exponential and polynomial functions?
A student applies L'Hospital's Rule to $\lim_{x \to 0} \frac{x^2 + 1}{x}$ and gets:
$$\lim_{x \to 0} \frac{x^2 + 1}{x} = \lim_{x \to 0} \frac{2x}{1} = 0$$
What is wrong with this reasoning?
Explain why L'Hospital's Rule uses $\frac{f'(x)}{g'(x)}$ rather than $\left(\frac{f(x)}{g(x)}\right)'$.
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.
Looking back:
Looking ahead:
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|---|---|---|
| Recognizing Indeterminate Forms | Skills Index | Converting Indeterminate Products |
Last updated: 2026-01-22