Math Core

Some limits can't be found by substitution because they produce a meaningless expression like $\dfrac{0}{0}$ . In Unit 1 you handled these with algebra: factoring, rationalizing, or special trig limits. L'Hôpital's rule gives a single, powerful method that uses derivatives instead, and it works on many limits where algebra gets stuck.

Indeterminate forms

Try substituting $x = 0$ into $\dfrac{\sin(3x)}{x}$ . You get $\dfrac{0}{0}$ . That doesn't mean the limit is 0, or 1, or undefined; it means substitution gave you no information. The numerator and denominator are both shrinking toward 0, and the limit depends on how fast each one shrinks.

The same thing happens with $\dfrac{\infty}{\infty}$ : in $\lim\limits_{x \to \infty} \dfrac{x^2}{e^x}$ , both parts grow without bound, and you need to know which grows faster. These two expressions are called indeterminate forms.

The rule

L'Hôpital's rule

Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$ ), $g'(x) \ne 0$ near $a$ , and

\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0 \quad\text{or}\quad \lim_{x \to a} f(x) = \pm\infty \text{ and } \lim_{x \to a} g(x) = \pm\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 $\pm\infty$ ). The same rule holds for one-sided limits and for $x \to \pm\infty$ .

Why does this work? When $f(a) = g(a) = 0$ , local linearity says $f(x) \approx f'(a)(x - a)$ and $g(x) \approx g'(a)(x - a)$ near $a$ . The factors of $(x - a)$ cancel, leaving $\dfrac{f'(a)}{g'(a)}$ . The ratio of two quantities that both vanish is decided by the ratio of their rates.

Common mistake

Check the form first, every time. L'Hôpital's rule applies only to $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ . For example, $\lim\limits_{x \to 0} \dfrac{x + 1}{x + 2} = \dfrac{1}{2}$ by substitution, but blindly differentiating top and bottom gives $\dfrac{1}{1} = 1$ , which is wrong. On the AP exam you must show that the form is indeterminate (by writing the two separate limits) to earn credit for using the rule.

Also: differentiate the numerator and denominator separately. This is not the quotient rule.

Worked examples

y = sin(3x)/x is undefined at x = 0, but the graph heads toward a height of 3.Open in grapher →

Worked example: A basic 0/0 limit

Evaluate $\lim\limits_{x \to 0} \dfrac{\sin(3x)}{x}$ .

Solution. As $x \to 0$ , $\sin(3x) \to 0$ and $x \to 0$ , so the form is $\dfrac{0}{0}$ . By L'Hôpital's rule,

\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{3\cos(3x)}{1} = 3.

Worked example: Applying the rule twice

Evaluate $\lim\limits_{x \to 0} \dfrac{e^x - 1 - x}{x^2}$ .

Solution. Substituting gives $\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}$ . Apply the rule:

\lim_{x \to 0} \frac{e^x - 1}{2x}.

This is still $\dfrac{0}{0}$ , so apply the rule again:

\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}.

Each time, confirm the new limit is still indeterminate before differentiating again.

Worked example: An infinity-over-infinity limit

Evaluate $\lim\limits_{x \to \infty} \dfrac{x^2}{e^x}$ .

Solution. Both numerator and denominator grow without bound: $\dfrac{\infty}{\infty}$ . Apply the rule twice:

\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.

Exponential functions eventually outgrow every power of $x$ . In the same way, $\lim\limits_{x \to \infty} \dfrac{\ln x}{x} = \lim\limits_{x \to \infty} \dfrac{1/x}{1} = 0$ : logarithms grow more slowly than any power.

Worked example: AP style: using given values

Functions $f$ and $g$ are differentiable with $f(2) = 0$ , $g(2) = 0$ , $f'(2) = 5$ and $g'(2) = -2$ , and $f'$ and $g'$ are continuous. Find $\lim\limits_{x \to 2} \dfrac{f(x)}{g(x)}$ .

Solution. Because $f$ and $g$ are continuous, $\lim\limits_{x \to 2} f(x) = f(2) = 0$ and $\lim\limits_{x \to 2} g(x) = 0$ , so the form is $\dfrac{0}{0}$ . Since $f'$ and $g'$ are continuous and $g'(2) \ne 0$ ,

\lim_{x \to 2} \frac{f(x)}{g(x)} = \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{f'(2)}{g'(2)} = \frac{5}{-2} = -\frac{5}{2}.

When to reach for the rule, and when not to

L'Hôpital's rule is a tool, not a reflex. A few guidelines keep it working for you:

Try substitution first. If substitution gives a real number, or a nonzero number over 0, the form is not indeterminate and the rule does not apply. A nonzero number over 0 points to an infinite limit or a vertical asymptote, which you analyze with signs, as in Unit 1.
Simplify between steps. After differentiating, cancel common factors or rewrite trig expressions before deciding whether to differentiate again. Messy expressions tend to get messier.
Watch for circles. Some limits, such as $\lim\limits_{x \to \infty} \dfrac{\sqrt{x^2 + 1}}{x}$ , bounce back and forth between two forms no matter how many times you apply the rule. When that happens, switch to algebra: dividing the numerator and denominator by $x$ gives the limit 1 immediately.
Algebra is still allowed. For a rational function like $\dfrac{x^2 - 9}{x - 3}$ , factoring is just as quick. Both methods earn full credit when done correctly.

The rule also sharpens what you learned about end behavior. Comparing growth rates with limits at infinity shows the ranking that AP questions rely on: logarithms grow more slowly than powers of $x$ , and powers grow more slowly than exponentials.

Tip

Many AP free-response questions give values of $f$ , $f'$ , $g$ and $g'$ in a table and ask for a limit of $\dfrac{f(x)}{g(x)}$ . Write the separate limits of the numerator and denominator, name the form, name the rule, then substitute. Those four short steps are worth the full point.