L'Hôpital's Rule
Apply L'Hôpital's Rule to evaluate indeterminate forms 0/0 and ∞/∞ by differentiating numerator and denominator separately. Convert other indeterminate forms (0·∞, ∞−∞, 0⁰, ∞⁰, 1^∞) to apply the rule.
A lesson of Calculus, in Applications of Derivatives, about 10 minutes.
By the end of this lesson, you will be able to apply L'Hôpital's Rule to resolve indeterminate forms $0/0$ and $\infty/\infty$, convert other indeterminate forms ($0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, $\infty^0$) for L'Hôpital's Rule, and know when not to apply it.
L'Hôpital's Rule
If $\lim_{x\to c}\dfrac{f(x)}{g(x)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, and the limit of $\dfrac{f'(x)}{g'(x)}$ exists, then:
$$\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}$$
Conditions: $f$ and $g$ must be differentiable near $c$ (not at $c$), and $g'(x) \neq 0$ near $c$.
Example 1 — $0/0$ Form
$$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\cos x}{1} = 1$$
Confirm the classic result with L'Hôpital.
Example 2 — $0/0$ Multiple Applications
$$\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$$
$x \to 0$: $e^0 - 1 - 0 = 0$, denominator $0$ → $0/0$.
Apply: $\dfrac{e^x - 1}{2x} \to 0/0$ again. Apply again: $\dfrac{e^x}{2} \to \dfrac{1}{2}$
Example 3 — $\infty/\infty$ Form
$$\lim_{x \to \infty}\frac{x^3}{e^x}$$
$\infty/\infty$. Apply L'Hôpital three times:
$\dfrac{3x^2}{e^x} \to \dfrac{6x}{e^x} \to \dfrac{6}{e^x} \to 0$
This confirms $e^x$ grows much faster than any polynomial.
Converting Other Indeterminate Forms
$0 \cdot \infty$: write as $\dfrac{0}{1/\infty}$ or $\dfrac{\infty}{1/0}$ to get $0/0$ or $\infty/\infty$.
$\infty - \infty$: combine into a single fraction, then apply.
$0^0, 1^\infty, \infty^0$: take $\ln$ → becomes $0\cdot\infty$ → convert → take $e^{\text{result}}$.
Example 4 — $0 \cdot \infty$ Form
$$\lim_{x \to 0^+}x\ln x$$
$= \lim_{x \to 0^+}\dfrac{\ln x}{1/x}$ → $\dfrac{-\infty}{\infty}$; apply L'Hôpital:
$= \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+}(-x) = 0$
Example 5 — $1^\infty$ Form
$$\lim_{x \to 0}(1+x)^{1/x}$$
Let $L = $ the limit. $\ln L = \lim_{x\to 0}\dfrac{1}{x}\ln(1+x) = \lim_{x\to 0}\dfrac{\ln(1+x)}{x}$ → $0/0$.
L'Hôpital: $\dfrac{1/(1+x)}{1} \to 1$.
So $\ln L = 1$, meaning $L = e$. ✓ (Classic limit)
When NOT to Apply L'Hôpital's Rule
- When the limit is NOT indeterminate: $\lim_{x\to 0}x/1 = 0$ directly.
- When it creates a cycle: $\lim_{x\to\infty}e^x/e^x = 1$ — applying L'Hôpital gives the same form. Use direct simplification.
Summary
L'Hôpital's Rule: applies when limit is $0/0$ or $\infty/\infty$ → replace $f/g$ with $f'/g'$.