Application of a Limit of a multivariable functionHELP

[*Helix*]

Application of a Limit of a multivariable function..HELP!

Homework Statement

if f (x,y) = (xy^2)/(x+y^2).
prove that for every real number a there is a path along which f (x,y) will approache a, as (x,y) is approaching (0,0).

Homework Equations

f (x,y) = (xy^2)/(x+y^2)
A as a element of R
(x,y) approches a, as (x,y) approaches (0,0)

The Attempt at a Solution

Don't know where to begin, any help would be great!

 

[Billy Bob]

Suggestion: Rather than trying to get "a" to be the limit, just try taking limits along various curves. Try taking the limit along the line y=kx. Try it along the parabola y=kx^2. Neither of these actually will work, but with luck they will give you a good idea.

 

[Dick]

How about setting f(x,y)=a and solving for a path x(y) such that f(x(y),y)=a, no matter what the value of y as y->0?

 

[*Helix*]

So how does setting x(y) or y(x) help me in this equation? I'm assuming I don't have to prove the actual limit of the function do I? I'm assuming its going to be zero, just from using some polar equations, x=rcosθ and y=rsinθ , and letting the function approach r . I know that showing the limits along different paths will prove if the given limit exsists..but how do we define "all paths", and what does that say about the value a? can I make a = f(x,y) = to say z? this question isn't making sense

 

[Dick]

Look, the limit isn't zero. Try this. Set f(x,y)=2. Put y=(1/2), then x=(-2/7). Put y=(1/4) then x=(-2/31). Put y=(1/8) then x=(-2/127). Do you see what's happening? I'm finding (x,y) points closer and closer to (0,0) such that f(x,y)=2. Does that mean the limit is 2?

 

[Billy Bob]

Did you try my idea yet? It does work, and pretty easily, once you "guess" the right family of paths. My suggested paths should give you enough hints if you try them.

Edited to add: Ooops, maybe I was wrong. Better stick with Dick's idea.

 

[*Helix*]

y is decreasing towards zero from the positive side, and x is increasing towards zero from the negative side. for the limit to exist, we have to approach a value along all paths that lead to that value, correct?

 

[Dick]

y is decreasing towards zero from the positive side, and x is increasing towards zero from the negative side. for the limit to exist, we have to approach a value along all paths that lead to that value, correct?

You have to approach the SAME value, no matter how (x,y) approaches (0,0) for the limit of f(x,y) to exist. Any single path may or may not have a limit if f(x,y) itself doesn't have a limit.

 

[*Helix*]

how do I prove that with this paticular problem then?

 

[Dick]

how do I prove that with this paticular problem then?

You can't prove it because the limit of f(x,y) as (x,y) approaches (0,0) doesn't exist. You aren't supposed to do that. You are supposed to find a path along which f(x,y) approaches a. I'm trying to suggest you can find a path along which f(x,y)=a. I found some points approaching (0,0) where f(x,y)=2. What's the curve along which f(x,y)=2? Wouldn't that be a good path to choose for the case a=2?

 

[*Helix*]

so I took some values around zero of this function (N/D is @ zero)...and I see what you're saying..about the limit not exsisting:

-0.05 0 -0.05
0 N/D 0
0.05 0 0.05

but if we let f(x,y) = a = or equal say, z ... then we can show that the limit is a, for ANY a, as (x,y) is approching (0,0) ...I can see what you're saying now, but how to show that as a mathematical proof for ANY a? is where I'm confused.

 

[Dick]

so I took some values around zero of this function (N/D is @ zero)...and I see what you're saying..about the limit not exsisting:

-0.05 0 -0.05
0 N/D 0
0.05 0 0.05

but if we let f(x,y) = a = or equal say, z ... then we can show that the limit is a, for ANY a, as (x,y) is approching (0,0) ...I can see what you're saying now, but how to show that as a mathematical proof for ANY a? is where I'm confused.

Write f(x,y)=a and solve the equation for x. Just do it. What do you get?

 

[*Helix*]

I think I'm getting there...tell me if this makes sense to you:

if we let the paths be: x=y and y=x:

then the equations reduce to
f(x,y=x) = x/2
f(x=y,y) = y/2

taking the limits of these two equations
x-->a gives a/2
y-->a gives a/2
a/2=a/2 = a

then the limit is a along this path as f(x,y)--> (0,0)

 

[Dick]

I think I'm getting there...tell me if this makes sense to you:

if we let the paths be: x=y and y=x:

then the equations reduce to
f(x,y=x) = x/2
f(x=y,y) = y/2

taking the limits of these two equations
x-->a gives a/2
y-->a gives a/2
a/2=a/2 = a

then the limit is a along this path as f(x,y)--> (0,0)

Nope, no sense whatsoever. Why are you ignoring my advice?

 

[*Helix*]

how would you express this equation explicitly in x?

 

[Dick]

how would you express this equation explicitly in x?

What equation? Do you mean f(x,y)=a?? Write it out. It reduces to a linear equation to solve for x. Just TRY it.

 

[*Helix*]

2 = xy^2/(x^2+y^2)

is this what you're talking about? how do I isolate x?

 

[Dick]

2 = xy^2/(x^2+y^2)

is this what you're talking about? how do I isolate x?

No it's not. Your original problem was f(x,y)=xy^2/(x+y^2). Start by multiplying both sides by the denominator.