L'Hôpital's Rule is a powerful technique for evaluating limits that initially give indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Instead of factoring or rationalizing, you differentiate the numerator and denominator separately — then try the limit again. Repeat as needed.
🎯 In this section you will learn
- The statement of L'Hôpital's Rule
- Indeterminate forms: $\frac{0}{0}$ and $\frac{\infty}{\infty}$
- How to apply the rule to various limit problems
- Other indeterminate forms: $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, etc.
📌 The Rule
$$ \text{If } \lim_{x \to a} \frac{f(x)}{g(x)} \text{ is of the form } \frac{0}{0} \text{ or } \frac{\infty}{\infty}, \text{ then } $$
$$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$
provided the limit exists (or is $\pm\infty$)
💡 Important Conditions
- The rule applies only to indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
- You differentiate the numerator and denominator separately — not using quotient rule!
- If the new limit is still indeterminate, apply the rule again.
- The rule also works for $x \to \infty$ or $x \to -\infty$.
Find $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}$.
①
Check form
As $x \to 0$, $\sin x \to 0$ and $x \to 0$ → $\frac{0}{0}$
②
Apply L'Hôpital's Rule
$\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(x) = 1$
③
New limit
$\lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1$
Find $\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2}$.
①
First application
$\frac{0}{0}$ form. $\frac{d}{dx}(1-\cos x) = \sin x$, $\frac{d}{dx}(x^2) = 2x$
②
New limit
$\lim_{x \to 0} \frac{\sin x}{2x}$ → still $\frac{0}{0}$
③
Second application
$\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(2x) = 2$
④
Result
$\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$
Find $\displaystyle \lim_{x \to \infty} \frac{e^x}{x^2}$.
- As $x \to \infty$, $e^x \to \infty$ and $x^2 \to \infty$ → $\frac{\infty}{\infty}$
- Apply L'Hôpital: $\frac{d}{dx}e^x = e^x$, $\frac{d}{dx}x^2 = 2x$ → $\lim_{x \to \infty} \frac{e^x}{2x}$ (still $\frac{\infty}{\infty}$)
- Apply again: $\frac{d}{dx}e^x = e^x$, $\frac{d}{dx}2x = 2$ → $\lim_{x \to \infty} \frac{e^x}{2} = \infty$
Find $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x}$.
- As $x \to \infty$, $\ln x \to \infty$, $x \to \infty$ → $\frac{\infty}{\infty}$
- Apply L'Hôpital: $\frac{d}{dx}\ln x = \frac{1}{x}$, $\frac{d}{dx}x = 1$
- $\lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$
📌 Other Indeterminate Forms
Sometimes limits look like $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, or $\infty^0$. You can often rewrite them as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ to apply L'Hôpital's Rule.
| Indeterminate Form | Rewrite | Method |
|---|
| $0 \cdot \infty$ | $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$ | Rewrite as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ |
| $\infty - \infty$ | Common denominator or factor | Rewrite as $\frac{0}{0}$ |
| $0^0$, $1^\infty$, $\infty^0$ | Take $\ln$ of both sides | Use $y = \lim f(x)^{g(x)}$, $\ln y = \lim g(x)\ln f(x)$ |
Find $\displaystyle \lim_{x \to 0^+} x \ln x$.
①
Identify form
As $x \to 0^+$, $x \to 0$, $\ln x \to -\infty$ → $0 \cdot (-\infty)$
②
Rewrite as $\frac{\infty}{\infty}$
$x \ln x = \frac{\ln x}{1/x}$
③
Apply L'Hôpital
$\frac{d}{dx}\ln x = \frac{1}{x}$, $\frac{d}{dx}(1/x) = -1/x^2$
④
Simplify
$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$
Find $\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$.
①
Let $y = \left(1 + \frac{1}{x}\right)^x$
Take $\ln$: $\ln y = x \ln\left(1 + \frac{1}{x}\right)$
②
Rewrite as $0 \cdot \infty$
As $x \to \infty$, $\ln(1+1/x) \to 0$, so $\ln y \to \infty \cdot 0$
③
Rewrite as $\frac{0}{0}$
$\ln y = \frac{\ln(1+1/x)}{1/x}$
④
Apply L'Hôpital
$\frac{d}{dx}\ln(1+1/x) = \frac{-1/x^2}{1+1/x} = \frac{-1}{x^2 + x}$, $\frac{d}{dx}(1/x) = -1/x^2$
⑤
Simplify
$\lim_{x \to \infty} \frac{-1/(x^2+x)}{-1/x^2} = \lim_{x \to \infty} \frac{x^2}{x^2 + x} = 1$
⑥
Conclusion
$\ln y \to 1$ ⇒ $y \to e$
⚠️ Common Mistakes
- Using L'Hôpital when the form is not indeterminate — if direct substitution works, just do it.
- Applying quotient rule instead of differentiating numerator and denominator separately — L'Hôpital uses $f'/g'$, not $(f/g)'$.
- Forgetting to check conditions — L'Hôpital requires $f$ and $g$ to be differentiable near $a$ and $g'(x) \neq 0$.
- Not simplifying before applying — sometimes simplifying first makes the rule unnecessary or easier.
🔍 Key Takeaways
- L'Hôpital's Rule applies to limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
- Differentiate numerator and denominator separately, then take the limit.
- If the result is still indeterminate, apply the rule again.
- For other indeterminate forms ($0\cdot\infty$, $\infty-\infty$, $1^\infty$, etc.), rewrite them as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first.
- L'Hôpital's Rule is a powerful tool, but always check if direct substitution or algebraic simplification works first.
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