L'Hôpital's Rule

L'Hospital's Rule

Part 1 of 5 — Indeterminate Forms and the Rule

Topic Overview

When Direct Substitution Gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Substituting into $\lim_{x \to a} \frac{f(x)}{g(x)}$ can give:

On the AP exam, L'Hospital's Rule is used for exactly these two indeterminate forms, $\frac{0}{0}$ and $\frac{\infty}{\infty}$.

L'Hospital's Rule

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

$\lim_{x \to a} f(x) = 0 \text{ and } \lim_{x \to a} g(x) = 0 \quad \text{or} \quad \lim_{x \to a} f(x) = \pm\infty \text{ and } \lim_{x \to a} g(x) = \pm\infty.$

Then

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

provided the limit on the right exists (or is $\pm\infty$). The same holds for one-sided limits and for $x \to \pm\infty$.

Two Things to Check First

1. The form: both limits are $0$, or both are infinite. If not, the rule does not apply.
2. Differentiability: you will differentiate the numerator and the denominator.

Worked Example

$\lim_{x \to 0} \frac{\sin(3x)}{x}$

Check: $\sin(0) = 0$ and the denominator is $0$, so the form is $\frac{0}{0}$.

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

Key Fact: Differentiate the numerator and the denominator separately. This is NOT the quotient rule.

Check the form first 🎯

Classify before you compute. 🔍

Apply the rule. ✍️

Key Takeaways — Part 1

• L'Hospital's Rule is for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ only
• Check the form before using it
• Replace $\frac{f}{g}$ by $\frac{f'}{g'}$: two separate derivatives, never the quotient rule
• If substitution gives an ordinary number, that number is the limit

Up Next: Part 2 — $\frac{0}{0}$ limits, including using the rule more than once.

L'Hospital's Rule

Part 2 of 5 — $\frac{0}{0}$ Limits, Including Repeated Use

Worked Example 1: Logarithm

$\lim_{x \to 1} \frac{\ln x}{x - 1}$

Check: $\ln 1 = 0$ and $1 - 1 = 0$, so the form is $\frac{0}{0}$.

$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{1/x}{1} = 1$

Worked Example 2: Trig with a Shift

$\lim_{x \to \pi} \frac{\sin x}{x - \pi}$

Check: $\sin \pi = 0$ and $\pi - \pi = 0$: $\frac{0}{0}$.

$\lim_{x \to \pi} \frac{\sin x}{x - \pi} = \lim_{x \to \pi} \frac{\cos x}{1} = \cos \pi = -1$

Worked Example 3: Agreement with Factoring

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

The rule gives $\lim_{x \to 2} \frac{2x}{1} = 4$, the same answer as factoring: $\frac{(x-2)(x+2)}{x-2} = x + 2 \to 4$. Both methods are valid; use whichever is quicker.

Using the Rule More Than Once

After one application, substitute again. If the new limit is still $\frac{0}{0}$ (or $\frac{\infty}{\infty}$), you may apply the rule again.

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

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

Limits That Are Derivatives in Disguise

$\lim_{h \to 0} \frac{e^{2 + h} - e^2}{h}$ is $\frac{0}{0}$. Treating $h$ as the variable, the rule gives $\lim_{h \to 0} \frac{e^{2+h}}{1} = e^2$. You can also recognize it as the derivative of $e^x$ at $x = 2$.

Key Fact: Re-check the form before every application. Once substitution gives an ordinary number, stop.

Practice — 0/0 limits 🎯

Track the form at each step. 🔍

Apply the rule (maybe twice). ✍️

Key Takeaways — Part 2

• Verify $\frac{0}{0}$, then differentiate the top and the bottom separately
• Substitute again after each application; repeat only while the form is still indeterminate
• L'Hospital's Rule agrees with factoring and with "derivative in disguise" reasoning

Up Next: Part 3 — $\frac{\infty}{\infty}$ limits and growth rates.

L'Hospital's Rule

Part 3 of 5 — $\frac{\infty}{\infty}$ Limits and Growth Rates

Worked Example 1: $\ln x$ Against $x$

$\lim_{x \to \infty} \frac{\ln x}{x}$

Check: both $\ln x$ and $x$ grow without bound, so the form is $\frac{\infty}{\infty}$.

$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$

Worked Example 2: A Power Against $e^x$ (Twice)

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

$\frac{x^2}{e^x} \;\to\; \frac{2x}{e^x} \;\to\; \frac{2}{e^x} \;\to\; 0$

The first two quotients are both $\frac{\infty}{\infty}$, so the rule is used twice; the last one is $\frac{2}{\infty}$, which is $0$.

Worked Example 3: A One-Sided $\frac{\infty}{\infty}$ Limit

$\lim_{x \to 0^+} \frac{\ln x}{1/x}$

As $x \to 0^+$, $\ln x \to -\infty$ and $\frac{1}{x} \to +\infty$: an $\frac{\infty}{\infty}$ form (the signs may differ).

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

Growth Rates as $x \to \infty$

Repeated use of the rule shows a clear ranking:

$\ln x \;\ll\; x^p \; (p > 0) \;\ll\; e^x$

When Algebra Is Faster

For a rational function such as $\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - x}$, the rule works twice ($\frac{6x}{10x - 1}$ is still $\frac{\infty}{\infty}$, then $\frac{6}{10} = \frac{3}{5}$), but dividing by $x^2$ gives $\frac{3}{5}$ at once. Use L'Hospital's Rule when logarithms, exponentials or trig functions block the algebra.

Practice — ∞/∞ limits 🎯

Which grows faster? 🔍

Apply the rule. ✍️

Key Takeaways — Part 3

• $\frac{\infty}{\infty}$ (with any signs) is the second form the rule handles
• Growth ranking as $x \to \infty$: $\ln x \ll x^p \ll e^x$
• For rational functions, dividing by the highest power is usually faster

Up Next: Part 4 — limits from tables and graphs, and how to justify the rule.

L'Hospital's Rule

Part 4 of 5 — Limits from Tables and Graphs, and Justification

On the AP exam the functions are often given only by a table or a graph. The rule works the same way: check the form from the given values, then use the given derivatives.

Worked Example: A Table

$f$ is twice differentiable, with $f(3) = 0$ and $f'(3) = 6$. Find $\lim_{x \to 3} \frac{f(x)}{x^2 - 9}$.

Justification and computation:

• $f$ is differentiable, so it is continuous: $\lim_{x \to 3} f(x) = f(3) = 0$.
• $\lim_{x \to 3} (x^2 - 9) = 0$.
• The limit has the form $\frac{0}{0}$, so L'Hospital's Rule applies:

$\lim_{x \to 3} \frac{f(x)}{x^2 - 9} = \lim_{x \to 3} \frac{f'(x)}{2x} = \frac{f'(3)}{6} = \frac{6}{6} = 1$

Why "twice differentiable" matters: it makes $f'$ continuous, so $\lim_{x \to 3} f'(x) = f'(3)$ and you may substitute the table value.

Writing the Justification

Before using the rule in a free-response answer, state the limits of the numerator and the denominator:

"Since $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, L'Hospital's Rule applies, so $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$."

For an $\frac{\infty}{\infty}$ limit, state that both limits are infinite instead.

AP Tip: Writing "$\frac{0}{0}$" by itself does not show which limits you checked. Name both limits, then apply the rule.

Worked Example: A Graph

Near $x = 2$ the graphs of $f$ and $g$ above are straight lines, and both pass through $(2, 0)$, so $\lim_{x \to 2} f(x) = 0$ and $\lim_{x \to 2} g(x) = 0$.

The slopes are $f'(2) = 3$ (from $(2, 0)$ to $(3, 3)$) and $g'(2) = -1$ (from $(0, 2)$ to $(2, 0)$). Because the derivatives are constant near $2$,

$\lim_{x \to 2} \frac{f(x)}{g(x)} = \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{3}{-1} = -3.$

Tables and justification 🎯

Read the graph above. 🔍

Limit from given values. ✍️

Key Takeaways — Part 4

• From a table or graph, use the given values to show both limits are $0$ (or both infinite)
• Then use the given derivative values: $\frac{f'(a)}{g'(a)}$ when $f'$ and $g'$ are continuous
• In free response, name both limits before you apply the rule

Up Next: Part 5 — common traps and the final assessment.

L'Hospital's Rule

Part 5 of 5 — Common Traps and Final Assessment

Trap 1: Using the Rule When the Form Is Not Indeterminate

$\lim_{x \to 0} \frac{\cos x}{1 + x}$

Substitution gives $\frac{1}{1} = 1$, which is the limit. Differentiating anyway gives $\frac{-\sin x}{1} \to 0$, a wrong answer. The rule's conclusion is only guaranteed for $\frac{0}{0}$ and $\frac{\infty}{\infty}$.

Trap 2: Using the Quotient Rule

The rule uses $\frac{f'(x)}{g'(x)}$, two separate derivatives. The quotient rule computes $\left(\frac{f}{g}\right)'$, a different expression that does not give the limit.

Trap 3: Applying It Again After the Form Is Gone

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

Once: $\frac{\sin x}{2x + 1}$. Substitution now gives $\frac{0}{1} = 0$, so the limit is $0$. Applying the rule a second time would give $\frac{\cos x}{2} \to \frac{1}{2}$, which is wrong.

Trap 4: Skipping the Justification

On free response, state that the numerator and the denominator both approach $0$ (or both are infinite) before using the rule.

Final Assessment — Set 1 🎯

Final Assessment — Set 2 🎯

Spot the trap. 🔍

Final challenge. ✍️

L'Hospital's Rule — Complete! 🎓

Key Fact: Check the form, differentiate the top and the bottom separately, and re-check the form after every step.