L’Hospital’s Rule

L’Hospital’s Rule is a fundamental method in calculus, used to evaluate limits that result in indeterminate forms such as $\frac{0}{0}$$ \frac{0}{0} $ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $. This rule is essential in both AP Calculus AB and BC, allowing students to resolve complex limits by differentiating the numerator and denominator separately. Mastery of L’Hospital’s Rule is crucial for efficiently solving limit problems and is a key topic for success on the AP Calculus exams.

Free AP Calculus AB Practice Test

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Learning Objectives

For the AP Calculus AB and BC exams, you should learn to identify when a limit results in an indeterminate form, such as $\frac{0}{0}$$ \frac{0}{0} $ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $, and apply L’Hospital’s Rule to resolve these limits by differentiating the numerator and denominator. In AP Calculus BC, extend this understanding to more advanced applications, including repeated applications of the rule and its use with improper integrals and series. Mastery of these skills is essential for achieving high scores on the exam.

Introduction to L’Hospital’s Rule

L’Hospital’s Rule is a powerful tool in calculus used to evaluate limits of indeterminate forms. When you encounter limits that result in the forms $\frac{0}{0}$$ \frac{0}{0} $​ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $​, L’Hospital’s Rule provides a method to resolve these indeterminate forms by differentiating the numerator and the denominator separately.

The Indeterminate Forms

Before applying L’Hospital’s Rule, it’s important to recognize the situations where it is applicable. The rule is used to evaluate limits that produce the following indeterminate forms:

• $\frac{0}{0}$$ \frac{0}{0} $

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

There are other indeterminate forms like $0 \cdot \infty, \infty - \infty ), 0^0, \text{ and } \left( 1^\infty \right)$$ 0 \cdot \infty, \infty - \infty ), 0^0, \text{ and } \left( 1^\infty \right) $, but these typically need to be manipulated into one of the above forms before applying L’Hospital’s Rule.

Statement of L’Hospital’s Rule

Suppose you have a limit of the form:

$\lim_{x \to c} \frac{f(x)}{g(x)}$$\lim_{x \to c} \frac{f(x)}{g(x)} $

If the limit results in an indeterminate form $\frac{0}{0}$$ \frac{0}{0} $ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $, then:

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

provided the limit on the right-hand side exists or is infinite.

Conditions for Applying L’Hospital’s Rule

To correctly apply L’Hospital’s Rule, the following conditions must be met:

• The limit of $\frac{f(x)}{g(x)}$$ \frac{f(x)}{g(x)} $as x approaches c must yield an indeterminate form $\frac{0}{0}$$ \frac{0}{0} $​ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $​.

• The functions f(x) and g(x) must be differentiable near x=c.

• The derivative $\frac{f'(x)}{g'(x)}$$ \frac{f'(x)}{g'(x)} $​ must exist.

Step-by-Step Process to Apply L’Hospital’s Rule

• Check the form of the limit: First, plug in the value of x approaching the limit into the original limit $\frac{f(x)}{g(x)}$$ \frac{f(x)}{g(x)} $​. Determine if it results in $\frac{0}{0}$$ \frac{0}{0} $ or $\frac{\infty}{\infty}$$ \frac{\infty}{\infty} $​.

• Differentiate the numerator and denominator: If the limit is an indeterminate form, find the derivatives f′(x) and g′(x).

• Evaluate the limit: Compute the limit $\lim_{x \to c} \frac{f'(x)}{g'(x)}$$ \lim_{x \to c} \frac{f'(x)}{g'(x)} $.

• Repeat if necessary: If the limit still results in an indeterminate form after applying L’Hospital’s Rule, apply the rule again.

Examples

Example 1

Evaluate the limit $\lim_{x \to 0} \frac{\sin x}{x}$$ \lim_{x \to 0} \frac{\sin x}{x} $.

Since direct substitution into the limit gives the indeterminate form $\frac{0}{0}$$\frac{0}{0}$​, we can apply L'Hospital's Rule:

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

Example 2

Evaluate the limit $\lim_{x \to \infty} \frac{1}{x}$$ \lim_{x \to \infty} \frac{1}{x} $.

This limit does not immediately appear to need L'Hospital's Rule, but if we rewrite it as $\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{1}{x}$$ \lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{1}{x} $​, it’s clear that L'Hospital's Rule applies:

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

Example 3

Evaluate the limit $\lim_{x \to 0} \frac{e^x - 1}{x}$$ \lim_{x \to 0} \frac{e^x - 1}{x} $.​.

Direct substitution gives $\frac{0}{0}$$\frac{0}{0}$​, so we apply L'Hospital's Rule:

$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1$$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1$

Example 4

Evaluate the limit $\lim_{x \to \infty} \frac{\ln x}{x}$$ \lim_{x \to \infty} \frac{\ln x}{x} $​.

This limit also gives the indeterminate form $\frac{\infty}{\infty}$$\frac{\infty}{\infty}$​, so we use L'Hospital's Rule:

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

Example 5

Evaluate the limit $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$$ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} $​.

Substituting x = 1 into the expression gives $\frac{0}{0}$$\frac{0}{0}$​, so apply L'Hospital's Rule:

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

Multiple Choice Questions

Question 1

Evaluate the limit: $\lim_{x \to 0} \frac{e^x - 1}{x}$$ \lim_{x \to 0} \frac{e^x - 1}{x} $

A) 0
B) 1
C) e
D) ∞

Answer: B) 1

Explanation: Applying L'Hospital's Rule, we differentiate:

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

Question 2

Which of the following limits can be evaluated using L'Hospital's Rule?

A) $\lim_{x \to 0} \frac{\sin x}{x}$$ \lim_{x \to 0} \frac{\sin x}{x} $
B) $\lim_{x \to \infty} \frac{1}{x}$$ \lim_{x \to \infty} \frac{1}{x} $
C) $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$$ \lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
D) All of the above

Answer: D) All of the above

Explanation: All given limits lead to indeterminate forms, so L'Hospital's Rule applies to each.

Question 3

Evaluate the limit: $\lim_{x \to \infty} \frac{\ln x}{x}$$\lim_{x \to \infty} \frac{\ln x}{x} $

A) 0
B) 1
C) ∞
D) −∞

Answer: A) 0

Explanation: Using L'Hospital's Rule: $ \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 $