Using L'Hospital's Rule for Determining Limits of Indeterminate Forms - AP Calculus AB Study Guide

The Core Idea: Using L'Hospital's Rule for Determining Limits of Indeterminate Forms

In calculus, we often evaluate limits by direct substitution. However, this method can sometimes lead to expressions that are not clearly defined, such as $\frac{0}{0}$ or $\frac{\infty }{\infty }$. These are known as indeterminate forms because they do not, by themselves, indicate what the value of the limit is. The limit could be zero, a finite number, or it might not exist at all.

L'Hospital's Rule provides a powerful and direct method for resolving limits that result in these specific indeterminate forms. By using derivatives, the rule allows us to transform a complicated or indeterminate limit of a quotient into a new, simpler limit. It states that if the limit of a quotient of two functions results in $\frac{0}{0}$ or $\frac{\infty }{\infty }$, then the original limit is equal to the limit of the quotient of their derivatives, provided this new limit exists.

L'Hospital's Rule

L'Hospital's Rule is a theorem used to evaluate limits of indeterminate forms.

The Rule:

Suppose that ${\lim }_{x\to c}\frac{f(x)}{g(x)}$ produces an indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$.

If ${\lim }_{x\to c}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists, then:

$$\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

This rule applies for limits as $x\to c$, $x\to c^{+}$, $x\to c^{-}$, $x\to \infty$, or $x\to -\infty$.

Understanding the Conditions for L'Hospital's Rule

The application of L'Hospital's Rule is governed by strict conditions that must be verified before the rule can be used. Failure to check these conditions is a common source of error.

1. The Limit Must be a Quotient: The rule applies only to limits of the form ${\lim }_{x\to c}\frac{f(x)}{g(x)}$.

2. The Limit Must be Indeterminate: You must first attempt to evaluate the limit by direct substitution. L'Hospital's Rule can only be applied if this substitution results in one of two specific indeterminate forms:

   • $\frac{0}{0}$ (i.e., ${\lim }_{x\to c}f(x)=0$ and ${\lim }_{x\to c}g(x)=0$)

   • $\frac{\infty }{\infty }$ (i.e., ${\lim }_{x\to c}f(x)=\pm \infty$ and ${\lim }_{x\to c}g(x)=\pm \infty$)

3. Differentiate Separately, Do Not Use the Quotient Rule: The core of the rule is to take the derivative of the numerator and the derivative of the denominator independently. You are finding the limit of $\frac{f^{\prime }(x)}{g^{\prime }(x)}$, not the limit of the derivative of $\frac{f(x)}{g(x)}$.

4. The New Limit Must Exist: The rule is only valid if the limit of the quotient of the derivatives, ${\lim }_{x\to c}\frac{f^{\prime }(x)}{g^{\prime }(x)}$, exists (is a finite number) or is $\pm \infty$.

Core Concepts & Rules

• Direct substitution is the first step in evaluating limits. If this process results in $\frac{0}{0}$ or $\frac{\infty }{\infty }$, the limit is in an indeterminate form.

• L'Hospital's Rule is a method for resolving limits that are in the indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$.

• To apply the rule, you replace the original quotient of functions with a new quotient of their derivatives: $\frac{f(x)}{g(x)}$ is replaced by $\frac{f^{\prime }(x)}{g^{\prime }(x)}$.

• The limit of the original function is equal to the limit of this new function, provided the new limit exists.

• It is mandatory to confirm that the limit is an indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ before applying L'Hospital's Rule.

Step-by-Step Example 1: Basic Application

Problem: Evaluate the limit ${\lim }_{x\to 0}\frac{e^{2x}-1}{\sin (x)}$.

Step 1: Verify the Indeterminate Form

First, use direct substitution by plugging in $x=0$.

• Numerator: $e^{2(0)}-1=e^0-1=1-1=0$

• Denominator: $\sin (0)=0$

Since direct substitution results in $\frac{0}{0}$, the limit is in an indeterminate form and we can apply L'Hospital's Rule.

Step 2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator separately.

• Derivative of the numerator: $\frac{d}{dx}(e^{2x}-1)=2e^{2x}$

• Derivative of the denominator: $\frac{d}{dx}(\sin (x))=\cos (x)$

Now, set up the new limit:

$$\underset{x\to 0}{\lim }\frac{e^{2x}-1}{\sin (x)}=\underset{x\to 0}{\lim }\frac{2e^{2x}}{\cos (x)}$$

Step 3: Evaluate the New Limit

Use direct substitution on the new limit expression.

$$\underset{x\to 0}{\lim }\frac{2e^{2x}}{\cos (x)}=\frac{2e^{2(0)}}{\cos (0)}=\frac{2(1)}{1}=2$$

Step 4: State the Conclusion

Because the new limit exists, the original limit is equal to this value.

$$\underset{x\to 0}{\lim }\frac{e^{2x}-1}{\sin (x)}=2$$

Step-by-Step Example 2: Exam-Style Application

Problem: Evaluate the limit ${\lim }_{x\to \infty }\frac{3x^2-5x}{2x^2+7}$.

Step 1: Verify the Indeterminate Form

As $x\to \infty$, both the numerator and the denominator approach $\infty$.

• Numerator: ${\lim }_{x\to \infty }(3x^2-5x)=\infty$

• Denominator: ${\lim }_{x\to \infty }(2x^2+7)=\infty$

This is the indeterminate form $\frac{\infty }{\infty }$, so L'Hospital's Rule is applicable.

Step 2: Apply L'Hospital's Rule (First Time)

Differentiate the numerator and the denominator separately.

• Derivative of the numerator: $\frac{d}{dx}(3x^2-5x)=6x-5$

• Derivative of the denominator: $\frac{d}{dx}(2x^2+7)=4x$

The new limit is:

$$\underset{x\to \infty }{\lim }\frac{3x^2-5x}{2x^2+7}=\underset{x\to \infty }{\lim }\frac{6x-5}{4x}$$

Step 3: Re-evaluate the Limit

Check the new limit. As $x\to \infty$, both $6x-5$ and $4x$ approach $\infty$. This is still the indeterminate form $\frac{\infty }{\infty }$. Because the result is still indeterminate, we can apply L'Hospital's Rule a second time.

Step 4: Apply L'Hospital's Rule (Second Time)

Differentiate the new numerator and denominator.

• Derivative of the numerator: $\frac{d}{dx}(6x-5)=6$

• Derivative of the denominator: $\frac{d}{dx}(4x)=4$

The final limit is:

$$\underset{x\to \infty }{\lim }\frac{6x-5}{4x}=\underset{x\to \infty }{\lim }\frac{6}{4}$$

Step 5: Evaluate the Final Limit

The limit of a constant is the constant itself.

$$\underset{x\to \infty }{\lim }\frac{6}{4}=\frac{6}{4}=\frac{3}{2}$$

Step 6: State the Conclusion

$$\underset{x\to \infty }{\lim }\frac{3x^2-5x}{2x^2+7}=\frac{3}{2}$$

Using Your Calculator

L'Hospital's Rule is an analytical technique, and you must show the steps of verifying the indeterminate form and applying the rule on the exam. A calculator cannot perform these symbolic steps for you.

However, you can use your calculator to check your answer by investigating the limit numerically or graphically.

To check ${\lim }_{x\to 0}\frac{e^{2x}-1}{\sin (x)}$:

1. Graphically:

   • Press Y= and enter the function: Y1 = (e^(2X) - 1) / sin(X).

   • Press GRAPH. Use the ZOOM feature to get a good view around $x=0$.

   • Press TRACE and enter values very close to 0, such as $0.001$ and $-0.001$. Observe that the y-values are very close to 2.

2. Numerically (Table):

   • Press 2nd then TBLSET(Table Setup). Set TblStart to 0 and ΔTbl to a small number like $0.001$.

   • Press 2nd then TABLE. Scroll up and down. You will see that as X gets closer to 0 from both sides, the Y1 values get closer to 2.

This numerical and graphical evidence supports the analytical answer of 2, but it does not replace the work required to apply L'Hospital's Rule.

AP Exam Quick Hit

Common Question Types

• Direct Calculation (Multiple Choice): You will be given a limit that results in an indeterminate form and asked to find its value.

  • Example: Find ${\lim }_{x\to 1}\frac{\ln (x)}{x-1}$. (Answer: 1)

• Tabular Data (Free Response): You will be given a table with values for functions and their derivatives at a certain point and asked to evaluate a limit.

  • Example: Given a table where $f(2)=0$, $g(2)=0$, $f^{\prime }(2)=5$, and $g^{\prime }(2)=-3$, find ${\lim }_{x\to 2}\frac{f(x)}{g(x)}$. You would recognize the $\frac{0}{0}$ form and use the derivatives from the table: $\frac{f^{\prime }(2)}{g^{\prime }(2)}=\frac{5}{-3}$.

• Repeated Application (Multiple Choice or Free Response): A problem where the first application of L'Hospital's Rule still results in an indeterminate form, requiring a second application.

  • Example: Find ${\lim }_{x\to 0}\frac{1-\cos (x)}{x^2}$. (Answer: 1/2)

Common Mistakes

• Applying the Quotient Rule: A very common error is to differentiate the entire expression $\frac{f(x)}{g(x)}$ using the quotient rule. L'Hospital's Rule requires you to differentiate the numerator and denominator separately.

• Not Verifying the Indeterminate Form: Applying the rule to a limit that is not $\frac{0}{0}$ or $\frac{\infty }{\infty }$. Always perform direct substitution first. Applying the rule incorrectly can lead to the wrong answer. For example, ${\lim }_{x\to 0}\frac{x+1}{x+2}=\frac{1}{2}$ by substitution, but incorrect application of L'Hospital's Rule would give ${\lim }_{x\to 0}\frac{1}{1}=1$.

• Algebraic Errors in Differentiation: Simple mistakes when finding $f^{\prime }(x)$ or $g^{\prime }(x)$ (e.g., chain rule errors, power rule errors) will lead to an incorrect final limit.

• Stopping After One Application: In problems requiring multiple applications of the rule, students sometimes stop after the first step, even if the new limit is still indeterminate. Always re-evaluate the limit after each application of the rule.