Fundamental limit in two variables Can I write that  $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{u\to0}\frac{\sin(u)}{u}$$
and, hence, that $\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$? If so, why can I do it?
 A: Just use polar coordinates:
$$x=r\cos t\;\;,\;\;r\sin t$$
and observe that $\;(x,y)\to (0,0)\iff r\to 0\;$ , so your limit's simply
$$\lim_{r\to 0}\frac{\sin r^2}{r^2}\stackrel{\text{l'Hospital}}=\lim_{r\to 0}\frac{2r\cos r^2}{2r}=1$$
A: You can say that IF the second limit exists. But it's possible to have similar limits where that substitution is not valid. For instance, 
$$
\lim_{x \to 0} sign(x^2) = 1
$$
while
$$
\lim_{u \to 0} sign(u)
$$
does not exist. (Here "Sign" is the function that's $+1$ for positive arguments, $-1$ for negative ones, and $0$ for $0$.)
As for "why can you do it", if you write out the definition of limit in terms of epsilon and delta, you can see that things work out. I think you need to take a square root of delta (or perhaps square delta) somewhere, but ...
More generally, you have
$$
\lim_{x \to a} f \circ g(x) = \lim_{u \to L} f(u) 
$$
where 
$$
L = \lim_{x \to a} g(x)
$$
provided all three limits exist. I'm pretty sure that this is proved in Spivak's Calculus in some detail, but I don't have my copy here at home with me. 
By the way, it's possible for 
$$
\lim_{x \to a} f \circ g(x)
$$
to exist even if 
$$
\lim_{u \to L} f (u)
$$
does not. As an example, consider $g(x) = 0$, the constant function, and $f(x) = sign(x)$; and take the limit of $f \circ g(x)$ as $x \to 0$. 
The generalization to two variables doesn't introduce any really new complexity here, so understanding the 1-variable case is sufficient. There are, however, other things where the generalization to two variables does introduce new complexity, so don't get complacent based on my simple explanation!
A: Yes you can, but it requires to put yourself sufficiently around 0, to have the monotony of the Sinc function.
Suppose $\;(x,y)\;$ verify :  $\;U \leq$ $x^2 +y^2\leq 2*U\;$  
$\;U\;$ is supposed to be very small, and in particular smaller than $\;\frac\pi2\;$.
On $[0, \frac{\pi}{2}]$ , Sinc is decreasing , so :
If you call your two variables function f, you get : 
Sinc(2*U) $\leq$ f(x,y) = Sinc( $x^2 +y^2$) $\leq$ Sinc(U)
By definition here, (x,y) -> (0,0) if and only if U->0 
hence there is your equivalence, and the limit is 1
