Question

Here is an argument that \(\ln 3\) equals \(\infty-\infty\). Where does the argument go wrong ? Give reasons for your answer. \(\ln 3=\ln 1-\ln \frac{1}{3}\) \(=\lim _{b \rightarrow \infty} \ln \left(\frac{b-2}{b}\right)-\ln \frac{1}{3}\) \(=\lim _{b \rightarrow x}\left[\ln \frac{x-2}{x}\right]_{3}^{b}\) \(=\lim _{b \rightarrow \infty}[\ln (x-2)-\ln x]_{3}^{b}\) \(=\lim _{b \rightarrow \infty} \int_{3}^{b}\left(\frac{1}{x-2}-\frac{1}{x}\right) d x\) \(=\int_{3}^{\infty}\left(\frac{1}{x-2}-\frac{1}{x}\right) d x\) \(=\int_{3}^{\infty} \frac{1}{x-2} d x-\int_{3}^{\infty} \frac{1}{x} d x\) \(=\lim _{b \rightarrow \infty}[\ln (x-2)]_{3}^{b}-\lim _{b \rightarrow \infty}[\ln x]_{3}^{b}\) \(=\infty-\infty .\)

Short answer

Question: Identify the errors in the following argument and explain why they are incorrect: Argument: \(\ln3 =\ln1-\ln\frac{1}{3}\) \(=\lim_{b\rightarrow\infty}\ln\left(\frac{b-2}{b}\right)-\ln\frac{1}{3}\) \(=\lim_{b\rightarrow\infty}[\ln(x-2)]_{3}^{b}-\lim_{b\rightarrow\infty}[\ln x]_{3}^{b}\) \(=\infty-\infty\) Answer: The argument contains the following errors: 1. Misinterpretation of the logarithm: The correct representation of \(\ln3\) should be \(\ln3=\ln3^1=\ln (3^1)\), not as \(\ln1-\ln\frac{1}{3}\). 2. Incorrect application of the limit: The limit should be applied correctly to the natural logarithm function's argument, which must be a positive real number. 3. Invalid computation using infinity: Expressions of the form "∞−∞" are indeterminate forms and cannot be treated as ordinary arithmetic operations. To correctly analyze such expressions, Limit laws or L'hôpital's rule should be used instead of assuming a subtraction operation can be performed.

Step-by-step solution

Misinterpretation of the Logarithm#

The first error occurs at the following step: \(\ln3 =\ln1-\ln\frac{1}{3}\) The correct interpretation of the logarithm rule should be: \(\ln3=\ln3^1=\ln (3^1) \) So the initial presentation of \(\ln3\) is incorrect.

Incorrect Application of the Limit#

Another issue is the incorrect application of the limit in this step: \(=\lim_{b\rightarrow\infty}\ln\left(\frac{b-2}{b}\right)-\ln\frac{1}{3}\) The natural logarithm function's argument should be a positive real number, and the limit should be applied correctly:

Invalid Computation Using Infinity#

The final step incorrectly assumes that subtracting two infinite limits gives a meaningful result: \(=\lim_{b\rightarrow\infty}[\ln(x-2)]_{3}^{b}-\lim_{b\rightarrow\infty}[\ln x]_{3}^{b}\) \(=\infty-\infty\) In reality, expressions of the form "∞−∞" are considered indeterminate forms, which cannot be treated as an ordinary arithmetic operation. When analyzing this algebraic expression, it is essential to resolve it using Limit laws or L'hôpital's rule rather than simply assuming a subtraction operation can be performed. So, in summary, the argument goes wrong due to misinterpreting the logarithm rule, incorrectly applying the limit, and inappropriately performing arithmetic operations with infinity. These errors lead to an incorrect evaluation of the natural logarithm of 3, which is not equal to \(\infty-\infty\).

Key concepts

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln \), is a fundamental concept in calculus and it possesses unique properties that make it invaluable for solving a wide variety of mathematical problems. One of the key properties is that the logarithm of a product is equal to the sum of the logarithms of its factors: \[ \ln(ab) = \ln(a) + \ln(b) \.\] Similarly, the logarithm of a quotient is the difference of the logarithms: \[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \.\] Another important property is the logarithm of a power, which states: \[ \ln(a^b) = b\ln(a) \.\]
These properties allow for the simplification of complex expressions and are critical when working with exponential functions and limits. For example, if we take the natural logarithm properties into account, the expression for \( \ln(3) \) given in the original exercise should not decompose into an indeterminate form like \( \infty - \infty \), but rather be treated according to these rules.
Proper understanding and application of natural logarithm properties ensure that students do not fall into common pitfalls, such as misinterpreting expressions that involve logarithms, especially when handling limits involving infinity.

Limit Laws

Limit laws are the bread and butter of calculus, offering a set of tools that allow mathematicians to evaluate the behavior of functions as they approach a certain point or infinity. Some of the most basic laws include:

• The sum law, which states the limit of a sum is the sum of the limits.
• The product law, expressing that the limit of a product is the product of the limits.
• The quotient law, outlining that the limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.

When approaching an expression like \( \ln(3) \) from the exercise, it is critical to apply these laws correctly.
For instance, the expression involving the limit as \( b \to \infty \) in the exercise should not be interpreted as just subtracting one infinity from another, because \( \infty - \infty \) is an indeterminate form with no clear value. Instead, one would apply the limit laws to the individual components within the function to gain meaningful insight into its behavior. By adhering to these laws, the analysis becomes more structured and avoids incorrect simplifications that lead to indeterminate forms.

L'Hôpital's Rule

L'Hôpital's rule is a powerful technique in calculus used to evaluate the limits of indeterminate forms, such as \( 0/0 \) and \( \infty/\infty \). To apply L'Hôpital's rule, one must take the derivative of the numerator and denominator separately, and then re-evaluate the limit of the resulting fraction.
This rule can only be used when the original limit results in an indeterminate form and after confirming that the derivatives exist. It cannot be used when dealing with an indeterminate difference like \( \infty - \infty \), which is present in the initial exercise.
When faced with the indeterminate form of \( \infty - \infty \) as in the case of \( \ln3 \), it is essential to further manipulate the expression to convert it into a form suitable for L'Hôpital's rule, or another applicable method. This manipulation often involves algebraic simplification or transformation that leverages the properties of logarithms to avoid the pitfalls of direct subtraction of infinities. Correct application of L'Hôpital's rule can then lead to a clear and finite limit, avoiding any misinterpretation or misuse of infinity in calculus problems.