L'Hôpital's Rule and Indeterminate Forms

L'Hôpital's Rule

By the end of this lesson, you will be able to identify indeterminate forms and apply L'Hôpital's Rule to evaluate limits that produce 0/0 or ∞/∞.

The Problem: Indeterminate Forms

Some limits produce forms that are algebraically ambiguous:

• $\dfrac{0}{0}$: both numerator and denominator → 0
• $\dfrac{\infty}{\infty}$: both go to infinity
• $0 \cdot \infty$, $\infty - \infty$, $0^0$, $\infty^0$, $1^\infty$ — other indeterminate forms

For these, direct substitution gives an ambiguous answer (not just "undefined" — the limit may still exist).

L'Hôpital's Rule

If $\lim_{x\to a}\frac{f(x)}{g(x)}$ produces an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and if $f$ and $g$ are differentiable near $a$, 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 equals $\pm\infty$).

You differentiate the numerator and denominator separately — NOT the quotient as a whole.

Examples

Example 1: $\lim_{x\to 0}\dfrac{\sin x}{x}$

Check: $0/0$ ✓ → apply L'Hôpital's:

$\lim_{x\to 0}\dfrac{\cos x}{1} = \dfrac{1}{1} = 1$

(This confirms the special trig limit from earlier.)

Example 2: $\lim_{x\to 1}\dfrac{x^3 - 1}{x^2 - 1}$

Check: both → 0, so $0/0$ ✓

$\lim_{x\to 1}\dfrac{3x^2}{2x} = \dfrac{3}{2}$

(Could also factor: $\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{x^2+x+1}{x+1} \to \frac{3}{2}$. Both methods work.)

Example 3: $\lim_{x\to\infty}\dfrac{e^x}{x^2}$ ($\infty/\infty$)

Apply L'Hôpital's: $\lim_{x\to\infty}\dfrac{e^x}{2x}$ — still $\infty/\infty$.

Apply again: $\lim_{x\to\infty}\dfrac{e^x}{2} = \infty$

The limit is $+\infty$ — exponential growth beats polynomial growth.

Example 4: $\lim_{x\to\infty}\dfrac{\ln x}{x}$ ($\infty/\infty$)

Apply L'Hôpital's: $\lim_{x\to\infty}\dfrac{1/x}{1} = \lim_{x\to\infty}\dfrac{1}{x} = 0$

Logarithmic growth is slower than linear growth.

Converting Other Indeterminate Forms