Proving the Limit of 2x-2y2/(x2 - y2) Does Not Exist

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1. Homework Statement

Prove or disprove that the following limit exists

lim x->0 and y->0 of
2x-2y²/(x² - y²)


2. Homework Equations

0 < ((x-a)²+(y-b)²)^.5 < delta

|f(x,y)-L| < epsilon

The Attempt at a Solution

If I take the limit of x to approach 1 of f(x,1) then the limit is equal to 1
If I take the limit of y to approach 1 of f(1,y) then the limit is equal to 2

I can then say that the limit in question DNE. To prove this, can't I just say since there are 2 different L's we will get 2 different $$\delta$$ 's and because of these 2 $$\delta$$ 's we will be able to satisfy at least one $$\delta$$ but not always both?

Does this make sense?

Thanks!

 

[Mark44]

1. Homework Statement

Prove or disprove that the following limit exists

lim x->0 and y->0 of
2x-2y²/(x² - y²)

This isn't (2x-2y²)/(x² - y²), is it?

2. Homework Equations

0 < ((x-a)²+(y-b)²)^.5 < delta

|f(x,y)-L| < epsilon

The Attempt at a Solution

If I take the limit of x to approach 1 of f(x,1) then the limit is equal to 1
If I take the limit of y to approach 1 of f(1,y) then the limit is equal to 2

Why are you letting x and y approach 1? Your limit is as (x, y) --> (0, 0).

I can then say that the limit in question DNE. To prove this, can't I just say since there are 2 different L's we will get 2 different $$\delta$$ 's and because of these 2 $$\delta$$ 's we will be able to satisfy at least one $$\delta$$ but not always both?

Does this make sense?

Thanks!

 

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This isn't (2x-2y²)/(x² - y²), is it?

oh crap sorry about that. Yes

(2x-2y²)/(x² - y²)

 

[hover]

Wow I really balled this up. This should be

1. Homework Statement

Prove or disprove that the following limit exists

lim x->1 and y->1 of
(2^x-2y^2)/(x^2 - y^2)


2. Homework Equations

0 < ((x-a)^2+(y-b)^2)^.5 < delta

|f(x,y)-L| < epsilon

3. The Attempt at a Solution

If I take the limit of x to approach 1 of f(x,1) then the limit is equal to 1
If I take the limit of y to approach 1 of f(1,y) then the limit is equal to 2

I can then say that the limit in question DNE. To prove this, can't I just say since there are 2 different L's we will get 2 different $$\delta$$'s and because of these 2 $$\delta$$'s we will be able to satisfy at least one but not always both?

Does this make sense?

Thanks!

 

[Mark44]

If you can show that the limit is different for two different paths, then that's enough to say that the limit doesn't exist. You don't need to go on and prove anything more.

 

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Awesome!
Thanks!