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**NonCommAlg** you don't need polar coordinates ... if $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ is such that $\displaystyle \lim_{\bold{x}\to\bold{0}}f(\bold{x})=0,$ then $\displaystyle L=\lim_{\bold{x}\to\bold{0}} \frac{\sin(f(\bold{x}))}{f(\bold{x})}=1,$ where

$\displaystyle \bold{x}=(x_1, \ ... \ ,x_n), \ \bold{0}=(0, \ ... \ , 0).$ because you can just put $\displaystyle f(\bold{x})=u.$ then since $\displaystyle u \rightarrow 0$ as $\displaystyle \bold{x} \rightarrow \bold{0},$ we'll have that

$\displaystyle L=\lim_{u\to0} \frac{\sin u}{u} = 1.$