Limit of x^2y^2/(x^2+y^2) as (x,y) approaches (0,0)

[Design]

Homework Statement

Show that lim(x,y) -> (0,0) f(x,y) does not exist

Homework Equations

f(x,y) = x² + y / (x² + y²)^1/2

The Attempt at a Solution

Let x = rcosθ y=rsinθ

so f(x,y) = r²cos²θ + rsinθ / (r²cos²θ + r²sin²θ)^1/2

= r(rcos²θ + sinθ) / ( r² (cos²θ + sin²θ))^1/2

= rcos²θ + sinθ

Don't know how to proceed the reasoning from here outthank you

 

[gomunkul51]

Hello!
The simplest way to show a multivariate limit does not exist is to show that the partial limits do not converge to the same value, i.e. first you check what is the single variable limit if x is treated as a variable and y is treated as a constants and then check the single variable limit when y is the variable and x is the constant.
If the 2 limits are different, then the multivariate limit does not exist.
(the opposite is not always true, i.e. if the limits were the same it is still not sufficient to show that the multivariate limit exists).

P.S. please explain why you chose to check the limit by converting it to polar form? :)

 

[Design]

That is how the prof showed it in class, So i don't know any other way of solving it.

When you mean treating 1 of them as a constant do i just take lim x->0 of
x² + y (some constant) / (x² + y²)^1/2 and then using L'hopital's rule from there or what?

 

[Design]

New solution : (Tell me if this works)

Consider f(x,0) and f(0,y)

Notice that when y =0
x²/ (x²)^1/2 = x

Notice that when x=0
y/(y²)^1/2 = 1

Since the limits differ, therefore limit DNE.

 

[gomunkul51]

I think you got the idea (in your last post).

 

[Design]

How would I show it when the limits actually exist? say x²y²/x²+y² as as (x,y) -> (0,0)