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Chapter 14: Problem 8

Find the limits. $$\lim _{(x, y) \rightarrow(1,1)} \ln \left|1+x^{2} y^{2}\right|$$

Short Answer

Expert verified
The limit is \( \ln 2 \).

Step by step solution

01

Understanding the Problem

The problem is to find the limit \( \lim_{(x, y) \rightarrow (1, 1)} \ln |1 + x^2 y^2| \). This involves evaluating how the expression \( \ln |1 + x^2 y^2| \) behaves as \( (x, y) \) approaches \( (1, 1) \).
02

Substitute the Limit Point

Substitute \( x = 1 \) and \( y = 1 \) into the expression \( 1 + x^2 y^2 \): \[ 1 + (1)^2 (1)^2 = 1 + 1 = 2 \] Therefore, the expression inside the logarithm becomes \( 2 \).
03

Evaluate the Logarithmic Expression

Now evaluate the logarithmic function at this point: \[ \ln |2| = \ln 2 \] Since the natural logarithm of a positive number is defined, and the absolute value of \( 2 \) is \( 2 \), this simplifies to \( \ln 2 \).
04

Conclusion of the Limit Evaluation

The limit of the function \( \ln |1 + x^2 y^2| \) as \( (x, y) \) approaches \( (1, 1) \) is \( \ln 2 \). Thus, the limit exists and is finite.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm, often denoted as \( \ln \), is a mathematical function with significant applications in calculus and mathematical analysis. It is the inverse of the exponential function with base \( e \), where \( e \approx 2.71828 \). This means for a positive number \( x \), the natural logarithm \( \ln(x) \) yields the power to which \( e \) must be raised to get \( x \). Natural logarithms have particular properties that make them useful in calculus:
  • \( \ln(1) = 0 \) because \( e^0 = 1 \).
  • \( \ln(ab) = \ln(a) + \ln(b) \) for any positive numbers \( a \) and \( b \).
  • \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
  • \( \ln(a^b) = b\ln(a) \).
Understanding these properties helps simplify complex logarithmic expressions, as seen when evaluating limits in multivariable calculus.
Limit Evaluation
Limit evaluation in calculus helps determine the behavior of functions as inputs approach specific values. To evaluate a limit, like \( \lim_{(x, y) \rightarrow (1, 1)} \ln |1 + x^2 y^2| \), the goal is to find the value the function approaches as \( (x, y) \) nears the point \((1, 1)\). First, substitute \( x = 1 \) and \( y = 1 \) into the expression to simplify it:
  • Compute \( 1 + x^2 y^2 \): substituting gives \( 1 + 1^2 \cdot 1^2 = 2 \).
  • The expression inside the logarithm is now \( \ln(2) \).
Since logarithms are continuous functions, if you can calculate a straightforward value without encountering indeterminate forms like \( 0/0 \), you can conclude that \( \ln(2) \) is the evaluated limit.
Multivariable Calculus
Multivariable calculus extends concepts from calculus to functions of several variables. Unlike single-variable calculus, we deal with limits, derivatives, and integrals in several dimensions. In the context of limit evaluation, consider the two-variable function \( f(x, y) = \ln|1 + x^2 y^2| \). To find limits like \( \lim_{(x, y) \rightarrow (a, b)} f(x, y) \), different approaches can be taken:
  • Direct Substitution: Substitute the points directly into the function if it results in a defined expression, as with the problem \( (1,1) \rightarrow \ln(2) \).
  • Path Testing: Evaluate the limit along different paths. This ensures the limit is consistent regardless of how \( (x, y) \) approaches \( (a, b) \).
  • Continuity Checks: If a function is continuous at a point, the limit can directly be taken as the function's value at that point.
These tools ensure that multivariable functions are handled accurately when determining limits, thereby helping in more complex mathematical modeling and problem-solving.

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Most popular questions from this chapter

In Exercises \(71-74,\) find a function \(z=f(x, y)\) whose partial derivatives are as given, or explain why this is impossible. $$\frac{\partial f}{\partial x}=3 x^{2} y^{2}-2 x, \quad \frac{\partial f}{\partial y}=2 x^{3} y+6 y$$

The condition \(\nabla f=\lambda \nabla g\) is not sufficient Although \(\nabla f=\lambda \nabla g\) is a necessary condition for the occurrence of an extreme value of \(f(x, y)\) subject to the conditions \(g(x, y)=0\) and \(\nabla g \neq \mathbf{0},\) it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of \(f(x, y)=x+y\) subject to the constraint that \(x y=16 .\) The method will identify the two points \((4,4)\) and \((-4,-4)\) as candidates for the location of extreme values. Yet the sum \(x+y\) has no maximum value on the hyperbola \(x y=16\) The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum \(f(x, y)=x+y\) becomes.

In Exercises \(71-74,\) find a function \(z=f(x, y)\) whose partial derivatives are as given, or explain why this is impossible. $$\frac{\partial f}{\partial x}=\frac{2 y}{(x+y)^{2}}, \quad \frac{\partial f}{\partial y}=\frac{2 x}{(x+y)^{2}}$$

Find two numbers \(a\) and \(b\) with \(a \leq b\) such that $$\int_{a}^{b}\left(6-x-x^{2}\right) d x$$ has its largest value.

Find the maxima, minima, and saddle points of \(f(x, y),\) if any given that $$ \begin{array}{l}{\text { a. } f_{x}=2 x-4 y \text { and } f_{y}=2 y-4 x} \\\ {\text { b. } f_{x}=2 x-2 \text { and } f_{y}=2 y-4} \\ {\text { c. } f_{x}=9 x^{2}-9 \text { and } f_{y}=2 y+4}\end{array} $$ Describe your reasoning in each case.
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