find $\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$ if $$f(x,y)=\frac{x^2y^2}{x^4+y^4}$$ is a 2 variable function, find $$\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$$
I really don't understand 2 variable limits. I understand that if limits from 2 or more paths aren't the same, the limit doesn't exist, but I don't know how to find the limit since there is an infinite number of paths possible.
 A: Approach #1: $\epsilon$-$\delta$ proof...probably too much work for the given problem. 
Approach #2: Convert to polar/spherical
Recalling that $x=r \cos \theta$ and $y= r \sin \theta$, we note that $(x,y) \to (0,0) \iff r \to 0$. As such,
$$\lim_{(x,y) \to (0,0) } \frac{ x^2 y^2}{x^4 + y^4} = \lim_{r \to 0} \frac{r^2 \cos^2 \theta \cdot r^2 \sin ^2 \theta}{r^4 ( \cos^4 \theta + \sin^4 \theta)}=\lim_{r \to 0} \frac{\cos^2 \theta \cdot  \sin ^2 \theta}{ \cos^4 \theta + \sin^4 \theta} =\frac{\cos^2 \theta \cdot  \sin ^2 \theta}{ \cos^4 \theta + \sin^4 \theta} $$
Since the last is expression is a non-constant function of $\theta$ (i.e., doesn't simplify to a constant using trig identities), this means that the limiting value depends on the angle taken to approach $(0,0)$, so there is no way to assign a single value to the limit along all possible paths.
Approach #3: lines
Let's look at what happens if we approach $(0,0)$ along lines $y=kx$:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^4 + y^4} = \lim_{(x, kx) \to (0,0)} \frac{x^2 (kx)^2}{x^4 + (kx)^4} = \lim_{x \to 0} \frac{k^2 x^4}{x^4 (1+k^4)} = \frac{k^2}{1+k^4}$$ 
Again, we got that the limiting value depended upon path taken to reach $(0,0)$, i.e., the slope of the particular line followed. Therefore, there is no consistent way to define what value $f(x,y)$ should approach near the origin.
Approach #4: Graph it
 
Notice that as we approach $(0,0)$ in different directions, we limit on different values between $0$ and $1/2$. There is no way to pick a specific value that all possible curves limit upon. 
A: For a function $f(x,y)$ that is not defined at $x=y=0,$ the def'n of $r=\lim_{x,y\to 0}f(x,y)$ is $$\forall d>0 \;  \exists e>0 \; \forall x,y \; (0<x^2+y^2<e^2\implies |f(x,y)-r|<d).$$ From the details in the answer by erfink, you can see that in this case there is no such $r.$
