Problem 54 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y, z) \rightarrow(0,1,0)} \ln e^{x z}(1+y)$$

Step-by-Step Solution

Answer

Answer: The limit of the function is 0.

Step 1: Simplify the given function

First, let's simplify the given function: $$ \ln e^{xz}(1+y) $$ We have two parts here: 1. \(\ln e^{xz}\) 2. \((1+y)\) Using the properties of the natural logarithm function, the first part simplifies to: $$ \ln e^{xz} = xz $$ Thus, the entire simplified function becomes: $$ f(x, y, z) = xz(1+y) $$

Step 2: Find the limit with the given values

As \((x,y,z)\) approaches \((0,1,0)\), we substitute these values into the simplified function: $$ \lim_{(x, y, z) \to (0, 1, 0)} xz(1+y) $$ Now, let's determine the limit: $$ \lim_{(x, y, z) \to (0, 1, 0)} xz(1+y) = (0)(0)(1+1) = 0 $$ The limit of the function \(f(x, y, z) = xz(1+y)\) as \((x, y, z) \to (0, 1, 0)\) is equal to 0.

Key Concepts

Natural Logarithm Properties

Understanding the properties of the natural logarithm is crucial in simplifying complex expressions. The natural logarithm, denoted as \( \ln \), has several key properties that can simplify calculus problems:

• \( \ln(e^x) = x \): This property is incredibly useful for simplification, as shown in the exercise. When the term \( \ln(e^{xz}) \) appears, it reduces directly to \( xz \).
• \( \ln(a \cdot b) = \ln a + \ln b \): This property helps break down products inside a logarithm into more manageable terms.
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \): Useful for dividing terms under a logarithm.

By applying these properties, you can transform seemingly complicated expressions into simpler, manageable forms. This is exactly what happens when we simplify \( \ln e^{xz}(1+y) \) to \( xz(1+y) \). Understanding these transformations makes evaluating limits much more straightforward.

Limit Evaluation

The process of evaluating limits in calculus is a method to understand how a function behaves as it approaches a particular point. This exercise deals with the multivariable limit of a function as the variables \((x, y, z)\) approach \((0,1,0)\). The steps are as follows:

• Simplify the function wherever possible, as seen with the natural logarithm properties. Here, the expression simplified to \( f(x, y, z) = xz(1+y) \).
• Substitute the values into the simplified function. For the point \((0, 1, 0)\), replace each variable accordingly.
• Calculate the expression: \( (0)(0)(1+1) = 0 \).

As the function approaches the specified point, if the outcome is consistent, you have successfully evaluated the limit. The overall approach helps to predict behavior or trends of a function near given points without needing to compute directly at those points.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of several variables. When evaluating limits in this context, it is important to consider how changes in each variable can independently affect the entire function. Key considerations in multivariable calculus include:

• Understanding the point the variables are approaching. For our exercise, this is \((0,1,0)\).
• Identifying simplifying opportunities using properties, such as breaking down the logarithm.
• Employing substitution to find the limit, taking careful steps to ensure every part of the expression is correctly approached.

Handling multivariable functions can be quite complex; however, breaking them into simpler parts, as seen with the given function \( f(x, y, z) = xz(1+y) \), allows for a more manageable approach. Learning to work with these kinds of functions enhances your understanding of how calculus operates in multi-dimensional scenarios.