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

[reddawg]

Homework Statement

By considering different paths, show that the given function has no limit
as (x,y) \rightarrow (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) \rightarrow (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 ?

 

[Mark44]

Homework Statement

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

 

[reddawg]

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.

 

[HallsofIvy]

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.

 

[Mark44]

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?

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.

 

[reddawg]

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.