Finding a Limit in two variables

[hsetennis]

Homework Statement

f(1/u,0)=1 and f(0,1/u)=-1 for all positive (integer) values of u. Prove whether or not the limit as (x,y) ->(0,0) exists.

Homework Equations

none.

The Attempt at a Solution

I argue that 0,0 is not in the domain of the function, but this neglects the behavior of f(x,y). So I feel like I'm missing a clue. It seems similar to the binary function: f(x)=-1 when x>0 and f(x)=-1 when x<0. But I'm unsure how to solve and prove the limit.

Edit:[ok, I'm saying that if (1/u) were to equal 0, then n would end up being infinity (not a positive integer), but again, this doesn't account for the -1/1 values of f(x,y)]

 

[algebrat]

Homework Statement

f(1/u,0)=1 and f(0,1/u)=-1 for all positive (integer) values of u. Prove whether or not the limit as (x,y) ->(0,0) exists.

Homework Equations

none.

The Attempt at a Solution

I argue that 0,0 is not in the domain of the function, but this neglects the behavior of f(x,y). So I feel like I'm missing a clue. It seems similar to the binary function: f(x)=-1 when x>0 and f(x)=-1 when x<0. But I'm unsure how to solve and prove the limit.

Edit:[ok, I'm saying that if (1/u) were to equal 0, then n would end up being infinity (not a positive integer), but again, this doesn't account for the -1/1 values of f(x,y)]

By definition of limit, x and y would never be equal to zero.

 

[hsetennis]

Why is that? Can you not take: {Lim as u→∞[f(1/u,0)]}=f(0,0)

I was thinking that if I evaluate: Lim_{u\rightarrow\infty}f(1/u,0)\neqLim_{u\rightarrow\infty}f(0,1/u)
Because 1≠-1

Any help is appreciated

 

[LCKurtz]

Why is that? Can you not take: {Lim as u→∞[f(1/u,0)]}=f(0,0)

You could only claim that if f(0,0) was defined, which it isn't, and if it were and it was actually equal to that limit.

I was thinking that if I evaluate: Lim_{u\rightarrow\infty}f(1/u,0)\neqLim_{u\rightarrow\infty}f(0,1/u)
Because 1≠-1

Any help is appreciated

You are getting close. Remember that for a two variable function, if$$
\lim_{(x,y)\rightarrow (a,b)} f(x,y)= L$$ that requires that you get $L$ as $(x,y)\rightarrow (a,b)$ along any path. What can you conclude from that?

 

[hsetennis]

I conclude that because a limit for a point must be equivalent from any real path, \lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1 via path f(x,y)=(1/u,0) but \lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1 via path f(x,y)=(0,1/u).

Is such a conclusion sound ?

 

[LCKurtz]

I conclude that because a limit for a point must be equivalent from any real path, \lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1 via path f(x,y)=(1/u,0) but \lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1 via path f(x,y)=(0,1/u).

Is such a conclusion sound ?

Your original question was whether or not the limit exists. What do you conclude about that?

 

[hsetennis]

I conclude that because a limit for a point must be equivalent from any real path and \lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1 via path f(x,y)=(1/u,0) but \lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1 via path f(x,y)=(0,1/u), the limit at (x,y)=(0,0) does not exist.

 

[algebrat]

By definition of limit, x and y would never be equal to zero.

I should have said x OR y are non zero.
That is, by definition of limit, (x,y)≠(0,0)

 

[LCKurtz]

I conclude that because a limit for a point must be equivalent from any real path and \lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1 via path f(x,y)=(1/u,0) but \lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1 via path f(x,y)=(0,1/u), the limit at (x,y)=(0,0) does not exist.

That's right.

 

[hsetennis]

Thanks for the help !