Chapter 6: Problem 15
Determine $$ \lim _{z \rightarrow 0} \frac{a^{z}-b^{z}}{c^{z}-d^{z}} $$ where \(a, b, c, d>0\) and \(c \neq d\).
Short Answer
Expert verified
\( \frac{\ln a - \ln b}{\ln c - \ln d} \)
Step by step solution
01
Understand the Indeterminate Form
We are asked to evaluate \( \lim_{z \rightarrow 0} \frac{a^{z} - b^{z}}{c^{z} - d^{z}} \). As \( z \to 0 \), each of \( a^{z}, b^{z}, c^{z}, \) and \( d^{z} \) approaches 1, since any positive number raised to the power of 0 is 1. This gives us an indeterminate form \( \frac{0}{0} \). Therefore, we need further evaluation to find the limit.
02
Apply L'Hôpital's Rule
Since the expression forms a \( \frac{0}{0} \) indeterminate form as \( z \to 0 \), L'Hôpital's Rule can be applied. This rule states that if \( \lim_{z \to c} \frac{f(z)}{g(z)} = \frac{0}{0} \), then \( \lim_{z \to c} \frac{f(z)}{g(z)} = \lim_{z \to c} \frac{f'(z)}{g'(z)} \), assuming the latter limit exists.
03
Differentiate the Numerator and Denominator
Differentiate the numerator \( a^{z} - b^{z} \) and the denominator \( c^{z} - d^{z} \) with respect to \( z \). The derivative of \( a^{z} \) with respect to \( z \) is \( a^{z} \ln a \), and similarly, \( b^{z} \) becomes \( b^{z} \ln b \). Therefore, the derivative of the numerator is \( a^{z}\ln a - b^{z}\ln b \). Similarly, the derivative of the denominator is \( c^{z}\ln c - d^{z}\ln d \).
04
Evaluate the New Limit
Apply the new derivatives in the limit: \( \lim _{z \rightarrow 0} \frac{a^{z}\ln a - b^{z}\ln b}{c^{z}\ln c - d^{z}\ln d} \). As \( z \rightarrow 0 \), this becomes \( \frac{1 \cdot \ln a - 1 \cdot \ln b}{1 \cdot \ln c - 1 \cdot \ln d} = \frac{\ln a - \ln b}{\ln c - \ln d} \).
05
Simplify the Result
The expression \( \frac{\ln a - \ln b}{\ln c - \ln d} \) is the final result after simplification. Since logarithms have properties such as \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \), this expression can also be written as \( \ln \left(\frac{a}{b}\right) / \ln \left(\frac{c}{d}\right) \), representing a logarithmic ratio.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus used to find limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you are faced with a limit where the direct substitution leads to one of these forms, you can use L'Hôpital's Rule to simplify the process of finding the limit.
Here's how it works:
For example, consider the limit \( \lim _{z \rightarrow 0} \frac{a^{z} - b^{z}}{c^{z} - d^{z}} \). As you direct substitute \( z = 0 \), the result becomes \( \frac{0}{0} \), an indeterminate form perfect for L'Hôpital's Rule to apply. You then differentiate both the numerator and the denominator to proceed with the evaluation.
Here's how it works:
- If \( \lim_{z \to c} \frac{f(z)}{g(z)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then you may differentiate the numerator and the denominator separately.
- Apply the limit to the new expression: \( \lim_{z \to c} \frac{f'(z)}{g'(z)} \).
- This usually simplifies the expression, allowing you to find a finite limit or a clearer form.
For example, consider the limit \( \lim _{z \rightarrow 0} \frac{a^{z} - b^{z}}{c^{z} - d^{z}} \). As you direct substitute \( z = 0 \), the result becomes \( \frac{0}{0} \), an indeterminate form perfect for L'Hôpital's Rule to apply. You then differentiate both the numerator and the denominator to proceed with the evaluation.
Indeterminate Forms
Evaluating a limit often leads us to expressions that are unclear or undefined, known as indeterminate forms. These typically include expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), among others. These forms indicate that the limit requires more work to find a meaningful result.
Indeterminate forms are signals that direct evaluation or substitution do not give conclusive answers.
In the exercise provided \( \lim _{z \rightarrow 0} \frac{a^{z} - b^{z}}{c^{z} - d^{z}} \), note how substituting \( z = 0 \) made each exponents turn into 1 resulting in \( a^z - b^z = 0 \) and \( c^z - d^z = 0 \). This results in the ambiguous \( \frac{0}{0} \) form.
To overcome such forms, mathematical techniques such as L'Hôpital's Rule or algebraic manipulation often come into play. These methods carve out a defined path to finding the actual limit or value.
Indeterminate forms are signals that direct evaluation or substitution do not give conclusive answers.
In the exercise provided \( \lim _{z \rightarrow 0} \frac{a^{z} - b^{z}}{c^{z} - d^{z}} \), note how substituting \( z = 0 \) made each exponents turn into 1 resulting in \( a^z - b^z = 0 \) and \( c^z - d^z = 0 \). This results in the ambiguous \( \frac{0}{0} \) form.
To overcome such forms, mathematical techniques such as L'Hôpital's Rule or algebraic manipulation often come into play. These methods carve out a defined path to finding the actual limit or value.
Logarithmic Functions
Logarithmic functions play a prominent role in solving advanced mathematical problems, including limits involving exponents.
A logarithmic function, represented as \( \ln \) for natural logarithms, is the inverse of an exponential function. Properties of logarithms include converting a difference into a ratio through the identity \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \).
In our specific problem, simplifying the final result using logarithms provides deeper insights. After applying L'Hôpital’s Rule and simplifying, the expression becomes \( \frac{\ln a - \ln b}{\ln c - \ln d} \), which, using the properties of logarithms, can be written as \( \ln \left(\frac{a}{b}\right) / \ln \left(\frac{c}{d}\right) \).
This simplification showcases the power of logarithms in breaking down expressions into simpler terms, making them easier to interpret and better understand.
A logarithmic function, represented as \( \ln \) for natural logarithms, is the inverse of an exponential function. Properties of logarithms include converting a difference into a ratio through the identity \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \).
In our specific problem, simplifying the final result using logarithms provides deeper insights. After applying L'Hôpital’s Rule and simplifying, the expression becomes \( \frac{\ln a - \ln b}{\ln c - \ln d} \), which, using the properties of logarithms, can be written as \( \ln \left(\frac{a}{b}\right) / \ln \left(\frac{c}{d}\right) \).
This simplification showcases the power of logarithms in breaking down expressions into simpler terms, making them easier to interpret and better understand.