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Chapter 12: Problem 1

Find the indicated limit or state that it does not exist. \(\lim _{(x, y) \rightarrow(1,3)}\left(3 x^{2} y-x y^{3}\right)\)

Short Answer

Expert verified
The limit is -18.

Step by step solution

01

Understand the Problem

We need to find the limit as the point \((x, y)\) approaches \((1, 3)\) for the expression \(3x^2y - xy^3\). This means evaluating how the function behaves near that point.
02

Substitute the Limit Point

Firstly, substitute \(x = 1\) and \(y = 3\) directly into the expression:\[3(1)^2(3) - (1)(3)^3\]
03

Calculate the Expression

Evaluate the substituted expression:\[3(1)^2 imes 3 = 9\]\[(1) imes (3)^3 = 27\]
04

Find the Difference

Subtract the second term from the first:\[9 - 27 = -18\]
05

Conclusion

Since the expression simplifies to a single numerical value when evaluated at the given point, the limit exists. The limit is \(-18\).

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Key Concepts

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

Limits of Functions
In multivariable calculus, limits of functions extend the idea of approaching a specific value from one-dimensional functions to two-dimensional functions. Consider a function of two variables, such as \( f(x, y) = 3x^2y - xy^3 \). The goal is to examine how this function behaves as the input approaches a certain point, like \((1, 3)\).
  • We begin by identifying a target point, which in this case is \((1, 3)\). This means we are interested in what value the function approaches as \(x\) gets closer to 1 and \(y\) gets closer to 3.
  • The process involves direct substitution: we replace \(x\) and \(y\) with their respective limiting values (1 and 3) in the function, simplifying the expression.

This approach is used to verify if a two-variable function can reach a defined value at a particular coordinate, thus determining if the limit exists.
Two-variable Calculus
Two-variable calculus deals with functions that have two independent variables. Unlike one-variable calculus, where functions deal with a single input variable, two-variable calculus involves points on a plane represented by coordinates \((x, y)\).In the given exercise:
  • We deal with the function \(f(x, y) = 3x^2y - xy^3\), which allows us to explore the relationship and interactions between two inputs.
  • This kind of function can be graphed in a 3-dimensional space, where the inputs \((x, y)\) define a plane and the output gives the function's height above the plane.
  • To find a limit in this context, it's crucial to understand how the function changes as the input approaches a specific point.

Two-variable calculus helps build a foundation for analyzing more complex multivariable systems, often involving terrain-like surfaces where the behavior at different points can vary significantly.
Continuous Functions
A continuous function in the realm of two-variable calculus maintains a consistent, unbroken surface, with no gaps or jumps. This continuous nature means that small changes in input lead to small changes in the output.
  • For a function like \( f(x, y) = 3x^2y - xy^3 \), it is deemed continuous at a point if the limit as \((x, y)\) approaches a specific point is equal to the function's output at that point.
  • When we substituted \( x = 1 \) and \( y = 3 \), and obtained a specific value \(-18\) after simplification, it showed that the function smoothly approaches this value without disruptions.
  • Continuity allows us to confidently state the behavior of a function at a point by merely looking at the surrounding values in its neighborhood.

This property is essential when considering real-world applications where smoothness and predictability are necessary, ensuring reliable performance even as conditions change slightly.

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Most popular questions from this chapter

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=e^{-x y} ; \mathbf{p}=(1,-1) ; \mathbf{a}=-\mathbf{i}+\sqrt{3} \mathbf{j}\)

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