Unit 4.7 – Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

AP® Calculus AB & BC | Turning Indeterminate Limits into Simple Calculations

L’Hospital’s Rule is a powerful technique for finding limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Instead of using algebraic manipulation, you can differentiate the numerator and denominator separately and re-evaluate the limit.

🔑 L’Hospital’s Rule: Main Formula

Classic Form

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \qquad\text{if the initial limit is of the form } \frac{0}{0}\text{ or }\frac{\infty}{\infty} \]

You can only apply this rule if the original limit results in one of these two indeterminate forms!

🚦 When You Can (and Can't) Use L’Hospital's Rule

Only for limits of quotients where direct substitution gives \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
Never use it for determinate forms or other types of limits like \(\frac{0}{c}, \frac{c}{0}, \frac{\infty}{c}, \frac{c}{\infty}\) where \(c \neq 0\).
If the new limit is still indeterminate, you can apply the rule again!
Other indeterminate forms like \(0\cdot \infty\), \(\infty-\infty\), \(0^0\), \(1^\infty\), or \(\infty^0\) must be algebraically rewritten into the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) before you can apply the rule.

📖 Examples: Step-by-step Application

Example 1: Basic \(\frac{0}{0}\) Form

\(\lim\limits_{x \to 2} \frac{x^2-4}{x-2}\)
Plugging in \(x = 2\) gives \(\frac{0}{0}\), so we can use L’Hospital’s Rule.
Differentiate the top and bottom separately: \(\frac{d}{dx}(x^2-4) = 2x\) and \(\frac{d}{dx}(x-2) = 1\).
The new limit is: \[\lim_{x \to 2} \frac{2x}{1} = 4\]

Example 2: Trigonometric Limit

\(\lim\limits_{x \to 0} \frac{\sin x}{x}\)
This is the indeterminate form \(\frac{0}{0}\). Applying L’Hospital’s Rule: \[\lim_{x\to 0} \frac{\cos x}{1} = \frac{\cos(0)}{1} = 1\]

Example 3 (After Rewrite): \(0 \cdot \infty\) Form

\(\lim\limits_{x \to 0^+} x \ln x\)
This is of the form \(0 \cdot (-\infty)\). Rewrite as a quotient: \(\lim\limits_{x \to 0^+} \frac{\ln x}{1/x}\).
This is now \(\frac{-\infty}{\infty}\), so we can apply L’Hospital’s Rule:
\[\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0\]

💡 Short Notes, Tricks & AP® Patterns

Always check for an indeterminate form before using the rule! Applying it incorrectly is a common mistake.
You can apply L’Hospital's Rule multiple times in a row, as long as the form remains \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
To handle \(\infty-\infty\), try to find a common denominator to turn it into a single quotient.
For exponential forms like \(0^0\), \(1^\infty\), or \(\infty^0\), let \(y\) be the limit, take the natural log of both sides (\(\ln y\)), evaluate the new limit, and then exponentiate the result to find \(y\).
The rule requires that the derivatives of the numerator and denominator exist near the limit point.

📝 Practice Problems

Try These Yourself:

\(\lim\limits_{x \to 0} \frac{e^x - 1}{x}\)
\(\lim\limits_{x \to \infty} \frac{2x^3 + 5x}{5x^3-2}\)
\(\lim\limits_{x \to 0^+} x^x\)

Answers (Setups):

This is \(\frac{0}{0}\). By L’Hospital's Rule: \(\lim\limits_{x \to 0} \frac{e^x}{1} = 1\).
This is \(\frac{\infty}{\infty}\). Applying the rule three times eventually gives \(\frac{12}{30}=\frac{2}{5}\). (Or, you can use the leading coefficients rule for polynomials).
This is \(0^0\). Let \(y=x^x\), so \(\ln y = x\ln x\). We found \(\lim\limits_{x \to 0^+} x\ln x = 0\). Since \(\ln y \to 0\), then \(y \to e^0 = 1\).

✏️ AP® Exam & FRQ Success for L’Hospital’s Rule

You must explicitly state that the limit is an indeterminate form (\(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)) to justify using L’Hospital’s Rule.
Show the differentiation step clearly (i.e., write the new limit with the derivatives).
For multiple applications of the rule, show each step and re-verify the indeterminate form.
Cite “L’Hospital’s Rule” by name in your FRQ justifications for full credit.