L'Hôpital's Rule

The Rule

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:

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

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

Indeterminate Forms

Forms where L'Hôpital's applies directly:

• $\frac{0}{0}$
• $\frac{\infty}{\infty}$

Other indeterminate forms (require rewriting):

• $0 \cdot \infty$ → rewrite as $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$
• $\infty - \infty$ → combine into single fraction
• $0^0$, $1^\infty$, $\infty^0$ → use logarithms

Important Notes

1. Only use when you have an indeterminate form
2. Take derivatives of numerator and denominator separately (not quotient rule!)
3. May need to apply multiple times
4. Works for $x \to a$, $x \to a^+$, $x \to a^-$, $x \to \infty$, $x \to -\infty$

Examples

Example 1: Basic 0/0

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = 1$

Example 2: Basic ∞/∞

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

Example 3: Multiple applications

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

$= \lim_{x \to 0} \frac{\sin(x)}{2x} \quad \text{(still } \frac{0}{0}\text{)}$

$= \lim_{x \to 0} \frac{\cos(x)}{2} = \frac{1}{2}$

Example 4: Form 0 · ∞

$\lim_{x \to 0^+} x \cdot \ln(x)$

Rewrite as:

$\lim_{x \to 0^+} \frac{\ln(x)}{1/x} \quad \text{(form } \frac{-\infty}{\infty}\text{)}$

$= \lim_{x \to 0^+} \frac{1/x}{-1/x^2}$

$= \lim_{x \to 0^+} (-x) = 0$

Example 5: Form ∞ - ∞

$\lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{1}{x}\right)$

$= \lim_{x \to 0} \frac{x - \sin x}{x \sin x} \quad \text{(form } \frac{0}{0}\text{)}$

$= \lim_{x \to 0} \frac{1 - \cos x}{\sin x + x \cos x} \quad \text{(still } \frac{0}{0}\text{)}$

$= \lim_{x \to 0} \frac{\sin x}{\cos x + \cos x - x \sin x}$

$= \frac{0}{1 + 1 - 0} = 0$

Example 6: Form 1^∞

$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$

Let $y = \left(1 + \frac{1}{x}\right)^x$, so $\ln y = x \cdot \ln\left(1 + \frac{1}{x}\right)$

$\lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \cdot \ln\left(1 + \frac{1}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{1/x}$

(form $\frac{0}{0}$)

$= \lim_{x \to \infty} \frac{\frac{-1/x^2}{1 + 1/x}}{-1/x^2}$

$= \lim_{x \to \infty} \frac{1}{1 + 1/x} = 1$

Therefore: $\lim_{x \to \infty} y = e^1 = e$

Example 7: Form 0^0

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

Let $y = x^x$, so $\ln y = x \ln x$

From Example 4: $\lim_{x \to 0^+} x \ln x = 0$

Therefore: $\lim_{x \to 0^+} y = e^0 = 1$

When L'Hôpital's Fails

Don't use L'Hôpital's when:

• The limit is not indeterminate (e.g., $\frac{1}{0} = \infty$, not indeterminate)
• You can simplify algebraically first (often easier)
• It leads to an infinite loop