How Can Different Paths Show No Limit Exists as (x,y) Approaches (0,0)?

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Homework Help Overview

The discussion revolves around evaluating the limit of the function f(x,y) = x4/(x4 + y4) as (x,y) approaches (0,0) by considering different paths. Participants are exploring whether the limit exists based on the behavior of the function along various paths.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting y=mx2 to analyze the limit, questioning the correctness of this substitution and its implications on the limit's existence. There are attempts to evaluate the limit along different paths and to understand how the value of m affects the outcome.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning and clarifying points of confusion. Some participants suggest that the limit's dependence on the path indicates that it does not exist, while others are verifying the correctness of the substitutions made.

Contextual Notes

There is a noted typo regarding the powers of y in the original problem statement, which has led to some confusion in the analysis. Participants are working through this clarification as they discuss the implications for the limit.

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Homework Statement


By considering different paths, show that the given function has no limit
as (x,y) [itex]\rightarrow[/itex] (0,0).

f(x,y) = x4/(x4 + y4)

Homework Equations





The Attempt at a Solution


My instructor taught me this process a while back and am unsure if it fits for this problem:

let y=mx2

limit as (x, mx2) [itex]\rightarrow[/itex] (0,0) of
x4/(x4 + mx4)

I tried this method seeing as letting x=0 yields limit = 0 and letting y = 0 yields limit = 1 ?
 
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reddawg said:

Homework Statement


By considering different paths, show that the given function has no limit
as (x,y) [itex]\rightarrow[/itex] (0,0).

f(x,y) = x4/(x4 + y4)

Homework Equations





The Attempt at a Solution


My instructor taught me this process a while back and am unsure if it fits for this problem:

let y=mx2

limit as (x, mx2) [itex]\rightarrow[/itex] (0,0) of
x4/(x4 + mx4)

I tried this method seeing as letting x=0 yields limit = 0 and letting y = 0 yields limit = 1 ?
Your substitution is incorrect. If y = mx2, then y4 ≠ mx4.
 
Oh right, it would become m2x4

Therefore it's x4/(x4 + m2x4)

Factoring out x4 yields 1/(1+m2)

In similar problems this method proved that the limit depended on the value of m, therefore it did not exist. Is this the case here? I would think so.
 
Yes, different values of that limit for different values of "m" mean that the limit depends upon the path. Recall that if a function has a limit then the value of f must be close to that limit. But this shows the value of f close to (0,0) on the line y=2x is different from the value on the line y= 3x.
 
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reddawg said:
Oh right, it would become m2x4
No, that's wrong as well.
If y = mx2 (as you have in the OP), then y4 = (mx2)4, right?
reddawg said:
Therefore it's x4/(x4 + m2x4)

Factoring out x4 yields 1/(1+m2)

In similar problems this method proved that the limit depended on the value of m, therefore it did not exist. Is this the case here? I would think so.
 
Mark44 said:
No, that's wrong as well.
If y = mx2 (as you have in the OP), then y4 = (mx2)4, right?


Actually, that was a typo on my part while identifying the problem statement. It is supposed to be
y2 not y4.

Thank you for taking the time to catch that mistake though.
 

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