/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Find the limits. $$\lim _{(x, ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Understand the Function

We have to find the limit of the function \( f(x,y) = \ln |1+x^2y^2| \) as \((x, y)\) approaches \((1, 1)\). Here, the expression inside the logarithm, \( |1 + x^2y^2| \), simplifies to \( 1 + x^2y^2 \) when it is positive.
02

Substitute the Values

Substitute \(x = 1\) and \(y = 1\) into the expression \(|1 + x^2y^2|\). This gives us \(|1 + 1^2 \times 1^2| = |1 + 1| = 2|\).
03

Evaluate the Natural Logarithm

Since \( |1 + x^2y^2| = 2 \) when \(x = 1\) and \(y = 1\), we evaluate \( \ln(2) \), which equals approximately 0.693.
04

Confirm Continuity

The function \( \ln|1 + x^2y^2| \) is continuous at \((1, 1)\) because the expression \(1 + x^2y^2\) is positive around this point. Therefore, the limit and the direct substitution result are valid.

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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 written as \( \ln(x) \), is a special type of logarithm with a base of \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It is frequently used in calculus and higher-level mathematics due to its natural appearance in growth processes and continuous compounding.
  • Unlike logarithms with other bases, the natural logarithm \( \ln(x) \) is the time needed to reach a certain level of continuous growth.
  • It is only defined for positive real numbers, meaning \( \ln(x) \) does not exist for \( x \leq 0 \).
  • The natural logarithm has a simple derivative: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \), making it particularly useful in differentiation.
In the context of our problem, the function \( \ln|1 + x^2y^2| \) represents the natural logarithm, ensuring that the expression inside of it must be positive for the logarithm to be defined. The value \( \ln(2) \), which arises when \((x, y)\) is \((1, 1)\), gives us approximately 0.693, indicating the logarithm's effectiveness in capturing the continuous growth rate at this point.
Continuity of Functions
Continuity is a fundamental concept in calculus, indicating that a function behaves "nicely" without sudden jumps or breaks. For a function to be continuous at a point \( x = a \), three conditions must be satisfied:
  • The function \( f(x) \) is defined at \( x = a \).
  • The limit \( \lim_{x \to a} f(x) \) exists.
  • The limit equals the function's value at that point: \( \lim_{x \to a} f(x) = f(a) \).
In our exercise, the function \( f(x,y) = \ln|1 + x^2y^2| \) is continuous at \((1, 1)\). This is because the expression \( 1 + x^2y^2 \) remains positive around \((1, 1)\), ensuring \( \ln|1 + x^2y^2| \) is defined and without any breaks or jumps near that point.
Hence, the limit using direct substitution is valid and trustworthy due to the function's continuous nature at \((1, 1)\).
Multivariable Functions
Multivariable functions, as the name implies, involve more than one input variable. Instead of mapping single numbers \( x \) to outputs \( y \), multivariable functions map pairs (or tuples) of numbers to an output value. For instance, in our problem, \[ f(x, y) = \ln|1 + x^2y^2| \]
  • This function takes two inputs \((x, y)\) and produces an output based on their combined effect.
  • Such functions can be visualized as surfaces in three-dimensional space, offering a visual context for understanding limits and continuity.
  • Evaluating limits for these functions means analyzing how they behave as the input point approaches a specific location.
In the step-by-step solution, substituting \( (x, y) = (1, 1) \) allows us to understand that the function effectively reduces the two variables into a single output. This demonstrates how multivariable calculus broadens our understanding and capabilities beyond single-variable scenarios by dealing with real-world applications that involve more complex interactions.

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