Is it possible for this function on $R^2$ to be defined continuously at $0$? I wont post all the functions.I am practicing on continuity.So i wanna use definitions but i am not really sure how? Suppose i have the function
$$f(x)= \frac{x^4-y^4}{(x^2+y^2)^2}$$ When $(x,y)\neq (0,0)$ and
$f(x)=b $ when $(x,y)=(0,0)$
Proof:To be defined continouesly at $0$ the limit at zero must exist.So there must exist the $$\lim_{(x,y)\to(0,0)}\frac{x^4-y^4}{(x^2+y^2)^2} $$ Or if i do a variables change to polar coordinates  $x=rcosθ $  $y=rsinθ$
 then$$\lim_{(r,θ)\to(0,0)} cos^2θ-sin^2θ $$ but that function does not depend on $r$  and so will give the same limit for All $r's$ since it must be unique?. So i said the function cannot be defined continuously .Is it right?Does not seem that rigorous to me i think i need a better explanation as to why it might not be defined and why when it is not depended on both variables it doesnt work.
Also i would like a proof using other  definitions(open sets of the preimage or  the ε,δ,) because i do not know how to use the other definitions except that a function is continuous to a point iff when x-->xo then f(x)-->f(xo).
 A: You have almost answered your own question...
When switching to polar coordinates, we do not need $\theta\rightarrow0$, in fact we must allow $\theta$ to do anything, as long as $r\rightarrow0$. This shows in fact that the function cannot be made continuous, since the limit will depend on $\theta$.
HINT an even easier approach is to consider the limit along the lines $y=mx$. Then the limit will depend on $m$...
A: Your computations appear correct with the exception that $(0,0)$ (the origin) is now just represented by $r=0$; you will also need to state that due to the limit being independent of $r$, you may approach from different angles and get different limits (e.g., approaching $r=0$ with $\theta=0$ gives a limit different than when $\theta=\pi/2$). This will prove that the function cannot be made continuous.
A: Already, $f$ is a continuous function from $\mathbb{R}^2 \setminus \{(0,0)\}$ to $\mathbb{R}$, and one might ask whether we can extend the domain of $f$ to all of $\mathbb{R}^2$ (that is, define $f(0,0)$) in such a way that $f$ is a continuous function from $\mathbb{R}^2$ to $\mathbb{R}$.  
Suppose $f$ could be extended in some way.  Then (a general property of continuous functions from $\mathbb{R}^n \rightarrow \mathbb{R}^m$), if $C$ is any point of $\mathbb{R}^2$, then the limit as $(x,y)$ goes to $C$ of $f(x,y)$ should exist and be equal to $f(C)$.  In particular, the limit as $(x,y)$ goes to $(0,0)$ should be whatever you defined $f(0,0)$ to be.  But you can show that the limit $$\lim\limits_{(x,y) \to (0,0)} \frac{x^4 - y^4}{(x^2 + y^2)^2}$$ does not exist at all (you can use the polar form, or stick to rectangular coordinates, and give two specific examples of paths going to $(0,0)$ for which you get different limiting values).  So no matter how you define $f(0,0)$, the resulting function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ will not be continuous.
A: Hint: try and factorize $x^4-y^4$.  
A: Consider $(x,y)=(t,0)$, then the function reduces to f(t)=1 so $\lim_{t\to 0}f(t)=1$; on the other hand, considering $(x,y)=(0,t)$, one gets $g(t)=-1$, $\lim_{t\to0}(t)=-1$. If the limit existed, it should have been unique and independent on how you approach the origin.
