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Magazine Chapter 14: Problem 6
Find the limit, if it exists, or show that the limit does not exist. $$ \lim _{(x, y) \rightarrow(2,-1)} \frac{x^{2} y+x y^{2}}{x^{2}-y^{2}} $$
Short Answer
Expert verified
The limit is \(-\frac{2}{3}\).
Step by step solution
01
Understanding the Problem
We are given a limit problem where we need to determine if the limit exists as \((x, y)\) approaches \((2, -1)\) for the function \(\frac{x^{2} y + x y^{2}}{x^{2} - y^{2}}\). This is a multivariable limit problem.
02
Substitute Simple Paths
First, solve the limit along the simple paths. Start with the path where \(y = -1\). Substitute \(y = -1\) into the expression:\[\frac{x^{2}(-1) + x(-1)^{2}}{x^{2} - (-1)^{2}} = \frac{-x^{2} + x}{x^{2} - 1}\]Taking the limit as \(x\) approaches 2 gives:\[\lim_{x \to 2} \frac{-x^{2} + x}{x^{2} - 1} = \frac{-2^2 + 2}{2^2 - 1} = \frac{-4 + 2}{4 - 1} = \frac{-2}{3}\]
03
Choose Another Path
Choose another path, e.g., approaching along \(x = 2\). Substitute \(x = 2\) into the expression:\[\frac{2^{2}y + 2y^{2}}{2^{2} - y^{2}} = \frac{4y + 2y^{2}}{4 - y^{2}}\]Take the limit as \(y\) approaches -1:\[\lim_{y \to -1} \frac{4y + 2y^{2}}{4 - y^{2}} = \frac{4(-1) + 2(-1)^{2}}{4 - (-1)^{2}} = \frac{-4 + 2}{4 - 1} = \frac{-2}{3}\]
04
Approach from a Diagonal Path
Consider a diagonal path such as \(y = x - 3\). Substitute \(y = x - 3\) into the expression and simplify:\[\frac{x^{2}(x-3) + x(x-3)^{2}}{x^{2} - (x-3)^{2}}\]Simplify and evaluate the limit as \(x\) approaches 2. You'll find that this path also leads to:\[\lim_{x \to 2} \frac{x^{2}y + xy^{2}}{x^{2} - y^{2}} = \frac{-2}{3}\]
05
Conclude the Existence of the Limit
Since multiple different paths including direct and diagonal paths give the same result \(\frac{-2}{3}\), the limit exists and is the same from any path approaching \((2, -1)\). Therefore, the limit is \(\frac{-2}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Two-variable calculus
Two-variable calculus is an extension of single-variable calculus to functions of two variables. Instead of examining how one variable affects a function, we consider how two independent variables impact it. This can make problems more complex, especially when dealing with limits, derivatives, and integrals.
- In multivariable calculus, a function is typically expressed as \( f(x, y) \), where both \( x \) and \( y \) are inputs.
- The function value can change depending on changes in either or both variables.
Limit existence
When approaching limits with two variables, the same fundamental ideas of limits in single-variable calculus apply, but with added complexity. Determining the existence of a limit in two-variable calculus involves checking if the limit along all possible paths yields a consistent result.
A limit exists if:
A limit exists if:
- The result is the same when evaluated along different paths approaching the same point.
- The function is approaching a specific value as \((x, y)\) nears the point.
Path analysis
Path analysis is an essential technique in determining the limit of a multivariable function. As you approach a point, you evaluate the limit along different paths to check for consistency in the results. This approach ensures that the found limit is genuine and not just a coincidental outcome from a single path.
Some common paths to test include:
Some common paths to test include:
- Horizontal path (keeping one variable constant)
- Vertical path (changing one variable while keeping the other constant)
- Diagonal paths (both variables change, such as \( y = x \) or \( y = x - 3 \))
Mathematical substitution
Mathematical substitution is a strategy used to simplify complex expressions by replacing variables with simpler forms. In two-variable calculus, substitutions can reveal a function's behavior as variables approach specific values.
For instance, in the given exercise, substitutions like \( y = -1 \), \( x = 2 \), or \( y = x - 3 \) help to evaluate the limit from different perspectives. By substituting these values, the expression becomes simpler, allowing for easier evaluation of the limit.
Utilizing substitution:
For instance, in the given exercise, substitutions like \( y = -1 \), \( x = 2 \), or \( y = x - 3 \) help to evaluate the limit from different perspectives. By substituting these values, the expression becomes simpler, allowing for easier evaluation of the limit.
Utilizing substitution:
- Reduces complexity by breaking down the expression into manageable parts.
- Allows for testing multiple paths to ensure limit consistency.
