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

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

When evaluating the limit of a quotient of functions, direct substitution can sometimes lead to an ambiguous or "indeterminate" form, such as $\frac{0}{0}$ or $\frac{\infty }{\infty }$. These forms do not have a defined value, and the limit cannot be determined without further analysis. L'Hospital's Rule provides a powerful method for resolving these specific indeterminate forms. The rule is based on the idea that if both the numerator and denominator of a fraction are approaching zero or infinity, the limit of the fraction is equivalent to the limit of the ratio of their rates of change (their derivatives).

This technique allows us to replace a complicated or indeterminate limit problem with a potentially simpler one involving the derivatives of the original functions. The rule can also be extended to handle other indeterminate forms, such as $1^{\infty }$, $0^0$, ${\infty }^0$, $\infty -\infty$, and $0\cdot \infty$, by first algebraically manipulating them into the required $\frac{0}{0}$ or $\frac{\infty }{\infty }$ quotient form.

Key Rules & Theorems

The primary theorem for this topic is L'Hospital's Rule.

L'Hospital's Rule

Let $f$ and $g$ be functions that are differentiable on an open interval containing $c$, except possibly at $c$ itself.

If the limit of the quotient $\frac{f(x)}{g(x)}$ as $x$ approaches $c$ produces an indeterminate form of the type $\frac{0}{0}$ or $\frac{\pm \infty }{\pm \infty }$, that is:

$$\underset{x\to c}{\lim }f(x)=\underset{x\to c}{\lim }g(x)=0$$

OR

$$\underset{x\to c}{\lim }f(x)=\pm \infty \text{and}\underset{x\to c}{\lim }g(x)=\pm \infty$$

Then, provided the limit on the right side exists (or is $\pm \infty$):

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

This rule also applies for one-sided limits ($x\to c^{+}$ or $x\to c^{-}$) and for limits as $x\to \pm \infty$.

Understanding Indeterminate Forms

L'Hospital's Rule can only be directly applied to limits of quotients that result in the forms $\frac{0}{0}$ or $\frac{\infty }{\infty }$. However, several other indeterminate forms exist. The key to solving limits with these forms is to use algebraic manipulation or logarithmic properties to convert them into a quotient that fits the criteria for L'Hospital's Rule.

The other indeterminate forms are:

• $1^{\infty }$

• $0^0$

• ${\infty }^0$

• $\infty -\infty$

• $0\cdot \infty$

Strategy for Conversion:

1. For the form $0\cdot \infty$:

   If you have a limit of the form ${\lim }_{x\to c}[f(x)\cdot g(x)]$ where $f(x)\to 0$ and $g(x)\to \infty$, you can rewrite the product as a quotient:

   $$f(x)\cdot g(x)=\frac{f(x)}{1/g(x)}\left(\text{which approaches }\frac{0}{0}\right)$$

   OR

   $$f(x)\cdot g(x)=\frac{g(x)}{1/f(x)}\left(\text{which approaches }\frac{\infty }{\infty }\right)$$

   Choose the form that results in a simpler derivative.

2. For the forms $1^{\infty }$, $0^0$, and ${\infty }^0$:

   These exponential forms are handled using logarithms. For a limit ${\lim }_{x\to c}[f(x){]}^{g(x)}$, follow these steps:

   a. Let $y=[f(x){]}^{g(x)}$.

   b. Take the natural logarithm of both sides: $\ln (y)=\ln ([f(x){]}^{g(x)})=g(x)\ln (f(x))$.

   c. Evaluate the limit of $\ln (y)$: $L={\lim }_{x\to c}[g(x)\ln (f(x))]$. This limit will now be in the form $0\cdot \infty$, which can be converted to a quotient as described above.

   d. Once you find the value $L$, the original limit is $e^L$. So, ${\lim }_{x\to c}y=e^L$.

3. For the form $\infty -\infty$:

   This form typically requires algebraic manipulation, such as finding a common denominator, factoring, or multiplying by a conjugate, to transform the expression into a single quotient.

Core Concepts & Rules

• Applicability: L'Hospital's Rule is exclusively for limits of quotients that evaluate to the indeterminate forms $\frac{0}{0}$ or $\frac{\infty }{\infty }$.

• Procedure: To apply the rule, you differentiate the numerator and the denominator separately and then take the limit of the new quotient. This is not the same as applying the quotient rule.

• Condition: 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 or is $\pm \infty$.

• Repeated Application: If, after applying L'Hospital's Rule, the resulting limit is still of the form $\frac{0}{0}$ or $\frac{\infty }{\infty }$, the rule can be applied again. This process can be repeated as many times as necessary until the limit is no longer indeterminate.

• Other Indeterminate Forms: The forms $1^{\infty }$, $0^0$, ${\infty }^0$, $\infty -\infty$, and $0\cdot \infty$ are also indeterminate but must be algebraically converted into a $\frac{0}{0}$ or $\frac{\infty }{\infty }$ quotient before L'Hospital's Rule can be used.

Step-by-Step Example 1: Basic Application

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

Step 1: Check the form of the limit.

Substitute $x=0$ into the numerator and denominator.

• Numerator: $e^{2(0)}-1=e^0-1=1-1=0$

• Denominator: $\sin (0)=0$

The limit is of the indeterminate form $\frac{0}{0}$. Therefore, the conditions for L'Hospital's Rule are met.

Step 2: Apply L'Hospital's Rule.

Differentiate the numerator and the denominator separately with respect to $x$.

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

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

Now, form the new limit with the derivatives:

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

Step 3: Evaluate the new limit.

Substitute $x=0$ into the new expression.

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

The limit exists and is equal to 2.

Final Answer:

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

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

Problem: Evaluate the limit $\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^{2x}$.

Step 1: Check the form of the limit.

As $x\to \infty$:

• The base, $\left(1+\frac{3}{x}\right)$, approaches $(1+0)=1$.

• The exponent, $2x$, approaches $\infty$.

The limit is of the indeterminate form $1^{\infty }$. This requires logarithmic manipulation before L'Hospital's Rule can be applied.

Step 2: Use logarithms to transform the expression.

Let $y={\left(1+\frac{3}{x}\right)}^{2x}$.

Take the natural logarithm of both sides:

$$\ln (y)=\ln \left({\left(1+\frac{3}{x}\right)}^{2x}\right)$$

$$\ln (y)=2x\cdot \ln \left(1+\frac{3}{x}\right)$$

Step 3: Evaluate the limit of the transformed expression.

We now need to find ${\lim }_{x\to \infty }\ln (y)={\lim }_{x\to \infty }\left[2x\cdot \ln \left(1+\frac{3}{x}\right)\right]$.

As $x\to \infty$, $2x\to \infty$ and $\ln (1+\frac{3}{x})\to \ln (1)=0$.

This is the indeterminate form $\infty \cdot 0$.

Step 4: Rewrite the expression as a quotient.

To apply L'Hospital's Rule, we must rewrite the product as a fraction. We can move the $2x$ term to the denominator as $\frac{1}{2x}$. A simpler choice is to move just the $x$ term.

$$\underset{x\to \infty }{\lim }\frac{2\ln \left(1+\frac{3}{x}\right)}{\frac{1}{x}}$$

Now, check the form again. As $x\to \infty$:

• Numerator: $2\ln (1+\frac{3}{x})\to 2\ln (1)=0$

• Denominator: $\frac{1}{x}\to 0$

The limit is now in the form $\frac{0}{0}$, so we can apply L'Hospital's Rule.

Step 5: Apply L'Hospital's Rule.

Differentiate the numerator and denominator separately.

• Numerator derivative: $\frac{d}{dx}\left[2\ln \left(1+\frac{3}{x}\right)\right]=2\cdot \frac{1}{1+\frac{3}{x}}\cdot \left(-\frac{3}{x^2}\right)=\frac{-6}{x^2(1+\frac{3}{x})}=\frac{-6}{x^2+3x}$

• Denominator derivative: $\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}$

The new limit is:

$$\underset{x\to \infty }{\lim }\frac{\frac{-6}{x^2+3x}}{-\frac{1}{x^2}}$$

Step 6: Simplify and evaluate the new limit.

Simplify the complex fraction:

$$\underset{x\to \infty }{\lim }\left(\frac{-6}{x^2+3x}\cdot \frac{-x^2}{1}\right)=\underset{x\to \infty }{\lim }\frac{6x^2}{x^2+3x}$$

This is still of the form $\frac{\infty }{\infty }$. We could apply L'Hospital's Rule again, or we can divide the numerator and denominator by the highest power of $x$, which is $x^2$.

$$\underset{x\to \infty }{\lim }\frac{\frac{6x^2}{x^2}}{\frac{x^2}{x^2}+\frac{3x}{x^2}}=\underset{x\to \infty }{\lim }\frac{6}{1+\frac{3}{x}}=\frac{6}{1+0}=6$$

So, we have found that ${\lim }_{x\to \infty }\ln (y)=6$.

Step 7: Exponentiate to find the original limit.

We found the limit of $\ln (y)$, not $y$. To find the original limit, we must solve for $y$.

Since ${\lim }_{x\to \infty }\ln (y)=6$, then ${\lim }_{x\to \infty }y=e^6$.

Final Answer:

$$\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^{2x}=e^6$$

Using Your Calculator

L'Hospital's Rule is an analytical technique; a calculator cannot perform the symbolic differentiation required. However, a graphing calculator is an excellent tool for verifying your answer.

To check the result of ${\lim }_{x\to c}\frac{f(x)}{g(x)}$:

1. Graphical Method:

   • Enter the function $Y_1=\frac{f(x)}{g(x)}$ into the graphing utility.

   • Graph the function, adjusting the window to be centered around the value $c$.

   • Use the TRACE feature to move the cursor along the curve as $x$ gets very close to $c$ from both the left and the right. Observe the corresponding $y$-value.

   • Alternatively, use the $CALC$ menu and select $1:value$. Enter a value extremely close to $c$ (e.g., if $c=2$, try $2.0001$ and $1.9999$). The output should be close to your calculated limit.

2. Table Method:

   • Enter the function $Y_1=\frac{f(x)}{g(x)}$.

   • Go to TBLSET (Table Setup). Set TblStart to $c$ and ΔTbl (delta table) to a very small number, like $0.001$.

   • View the TABLE. The $Y_1$ values for $x$ values near $c$ should be approaching your calculated limit.

For limits approaching infinity, you can use the same methods by entering very large positive or negative numbers for $x$.

AP Exam Quick Hit

Common Question Types

• Direct Application from Functions: You will be given a limit of a quotient of two functions and asked to evaluate it. This is common in the multiple-choice section.

  • Example: Find $\underset{x\to 2}{\lim }\frac{x^2-4}{\ln (x-1)}$.

• Application from a Table of Values: A free-response question may provide a table with values for functions $f$, $g$, and their derivatives $f^{\prime }$ and $g^{\prime }$ at a specific point. You will be asked to evaluate a limit using this information.

  • Example: Given $f(3)=0$, $g(3)=0$, $f^{\prime }(3)=6$, and $g^{\prime }(3)=2$, find $\underset{x\to 3}{\lim }\frac{f(x)}{g(x)}$.

  • Solution: Since ${\lim }_{x\to 3}f(x)=0$ and ${\lim }_{x\to 3}g(x)=0$, the limit is of the form $\frac{0}{0}$. By L'Hospital's Rule, ${\lim }_{x\to 3}\frac{f(x)}{g(x)}={\lim }_{x\to 3}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\frac{f^{\prime }(3)}{g^{\prime }(3)}=\frac{6}{2}=3$.

• Limits Requiring Manipulation: You will be given a limit in a form like $1^{\infty }$ or $0\cdot \infty$ that must first be algebraically converted into a $\frac{0}{0}$ or $\frac{\infty }{\infty }$ quotient before the rule can be applied.

  • Example: Find $\underset{x\to 0^{+}}{\lim }{x}^2\ln (x)$.

Common Mistakes

• Using the Quotient Rule: The most frequent error is applying the quotient rule to the fraction $\frac{f(x)}{g(x)}$. L'Hospital's Rule requires taking the derivatives of the numerator and denominator separately: $\frac{f^{\prime }(x)}{g^{\prime }(x)}$, not $\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}$.

• Applying the Rule to a Determinate Form: Students sometimes apply L'Hospital's Rule without first checking if the limit is an indeterminate form. If direct substitution yields a value like $\frac{1}{0}$ or $\frac{5}{3}$, the rule does not apply and will produce an incorrect answer. Always substitute first.

• Forgetting the Final Step for Logarithmic Forms: When solving limits of the form $1^{\infty }$, $0^0$, or ${\infty }^0$, it is common to find the limit of the logarithm, $L=\lim \ln (y)$, and forget to exponentiate to find the final answer, $e^L$.

• Stopping Too Early or Continuing Too Long: When repeated applications are necessary, make sure to check the form of the limit after each application. Stop applying the rule as soon as the limit is no longer indeterminate. Conversely, if the limit is still indeterminate, you must apply the rule again.

• Algebraic Errors: Mistakes in differentiation (especially with the chain rule) or in the algebraic manipulation required to convert indeterminate forms are common pitfalls. Double-check your algebra and derivatives carefully.