Limit of ((x^3 + (4x^2)y)/(x^2+2y^2)) as (x,y)->(0,0)

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Homework Help Overview

The discussion revolves around finding the limit of the expression \(\frac{x^3 + (4x^2)y}{x^2 + 2y^2}\) as \((x,y)\) approaches \((0,0)\). The subject area involves multivariable calculus and limit evaluation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine the limit and expresses uncertainty about the next steps after stating a guess that the limit is 0. Some participants suggest using polar coordinates as a method to simplify the evaluation, while others clarify the transformation needed.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on using polar coordinates. There is an ongoing exploration of the approach to take, and multiple interpretations of the limit's behavior are being considered.

Contextual Notes

The original poster expresses discomfort with formatting and a lack of familiarity with polar coordinates, which may affect their ability to proceed with the problem effectively.

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Homework Statement



Find the limit of ((x^3 + (4x^2)y)/(x^2+2y^2)) as (x,y) -> (0,0)

Homework Equations


The Attempt at a Solution



I am guessing the limit is equal to 0, and I know I have to use

0 < sqrt(x^2 + y^2) < delta where |f(x,y) - L| < epsilon

I just have no idea what to do next
I don't want anyone to give me the answer, just a point in the right direction

Sorry about the formatting, I'm quite lost when it comes to Latex.
 
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Don't worry about the formatting. We can all read that. You used parentheses and everything. Express it in polar coordinates.
 
Sorry it's been a while since I did anything to do with polar, how do I do that?
 
x=r*cos(theta), y=r*sin(theta). You get an r^3 in the numerator and an r^2 in the denominator, yes? Cancel the r^2. Now the limit is r->0.
 

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