Find values for which the limit exists

  • Thread starter Thread starter twoflower
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary

Homework Help Overview

The problem involves finding conditions for the variables a, b, and c such that the limit of a given function as (x, y) approaches (0, 0) exists. The function in question is a rational expression involving the variables and their products.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various approaches to determine the conditions under which the limit exists, including the initial observation that non-zero values for the variables lead to non-existence of the limit. Some participants suggest using polar coordinates as a method to analyze the limit more effectively.

Discussion Status

The discussion is ongoing, with participants exploring different methods and questioning the assumptions made about the variables. Some guidance has been offered regarding the use of polar coordinates, and a few conditions have been proposed, though there is no consensus on the exact conditions required for the limit to exist.

Contextual Notes

There is mention of potential constraints regarding the use of L'Hôpital's rule and the nature of the variables involved, as well as the implications of guessing methods in finding solutions.

twoflower
Messages
363
Reaction score
0
Hi,
I'm given this problem:

Find conditions for variables a, b, c so that the limit

[tex] \lim_{[x,y] \rightarrow [0,0]} \frac{xy}{ax^2 + bxy + cy^2}[/tex]

exists.

What I have only found so far is that for all variables non-zero the limit doesn't exist. Anyway, I have no clue how to find the conditions for which it does. I tried a = b = c = 0, but it doesn't seem to help to me...

Thank you for the enlightenment.
 
Physics news on Phys.org
OOps I was too late in deleting my post. Actually I made a mistake in solving that and yes, what makes me unsure is that I don't think we could use hopital rule for these kind of limit.
 
You'd better to ask another homework helper, but I solve it in another way now . If you look at numerator nad denominator, you see both of them have xy. So?
 
Lisa! said:
You'd better to ask another homework helper, but I solve it in another way now . If you look at numerator nad denominator, you see both of them have xy. So?

Ok, now it seems to me that the condition for the limit to exist is that a = c = 0.

Anyway, it is just the result of guessing method, is there any more exact approach to solve this?
 
For functions like this, where you have two variables, I find it best to convert to polar coordinates. That way, exactly one variable, r, measures the distance to (0,0) which is the crucial factor. In polar coordinates,
[itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so
[itex]xy= r^2 cos(\theta)sin(\theta)[/itex], [tex]x<sup>2</sup>= r^2 cos^2(\theta)[/itex], and [tex]y^2= r^2 sin^2(\theta)[/itex].<br /> Of course, then [itex]ax^2+ bxy+ cy^2= ar^2cos^2(\theta)+ br^2sin(<br /> theta)cos(\theta)+ cr^2sin^2(\theta)[/itex] so that<br /> [itex]ax^2+ bxy+ cy^2= r^2(acos^2(\theta)+ bsin(<br /> theta)cos(\theta)+ csin^2(\theta)[/itex].<br /> <br /> That means that<br /> [tex]\frac{xy}{ax^2+ bxy+ cy^2}= \frac{sin(\theta)cos(\theta)}{acos^2(\theta)+ bsin(\theta)cos(\theta)+ csin^2(\theta)}[/tex].<br /> <br /> Notice that there is no "r" in that! This can have a limit as r-> 0 only if it does NOT depend on [itex]\theta[/itex]- it is a constant. One obvious choice for a,b,c is a= c= 0, b= 1 but there may be others.[/tex][/tex]
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
35
Views
5K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K