solving $ \lim_{(x,y) \rightarrow (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} $ in solving $ \lim_{(x,y) \rightarrow (0,0)}  \frac{\sin(x^2+y^2)}{x^2+y^2} $, my textbook says to replace $x^2+y^2$ with $r^2$, thus making the equation $ \lim_{r \rightarrow 0}  \frac{\sin(r^2)}{r^2} $. I understand how to solve this; it becomes  $\lim_{r \rightarrow 0}  \frac{2r\cos(r^2)}{2r} $ , which is cos(0), which is 1. But i do not understand why $x^2+y^2$ is replaced with $r^2$. can it be just r? 
then the equation becomes $ \lim_{r \rightarrow 0}  \frac{\sin(r)}{r} $ which is 1... Why is $x^2+y^2$ replaced with $r^2$ , not just  $r$? what am i missing?
 A: You are just using polar coordinates: $x=r \cos \theta$, $y=r\sin \theta$, so that $x^2+y^2 \to 0$ becomes $r^2 \to 0$, i.e. $r \to 0$.
A: It usually makes more sense to do the substitution $r^2 = x^2+y^2$, since this corresponds to polar coordinates. Why? Because it is easy to visualize and to get the inverse transformation,
\begin{aligned}
x & = r\cos\theta\\
y & = r\sin\theta,
\end{aligned}
in case you would ever need it (e.g. for interpreting the result, additional manipulations, or transforming back to the original variables). Your proposed substitution $\rho = x^2+y^2$ works just as well:
$$\lim_{(x,y)\to(0,0)} = \frac{\sin(x^2+y^2)}{x^2+y^2} = \lim_{\rho\to0}\frac{\sin(\rho)}{\rho} = 1.$$
What it boils down to in this case is personal preference and tradition. If this was part of a bigger problem, or the problem was a bit more complex, the polar coordinate transformation would probably offer more versatility. For example, the trigonometry and the radius $r$ (without the square) would not be as well represented in the $\rho$ coordinate system.
