The requisite extension assigns to the origin the value of the bivariate limit of $f$ at the origin. Hence, what is required is to show that this limit is 0. Unfortunately, showing $\left|f-0\right|\to 0$ by the "usual" techniques of estimation turns out to be a significant challenge. So also is the approach taken in Example 3.2.25 where the maxima of $f$ on circles of radius $r$ are computed, and then shown to tend to $L$ as $r\to 0$. At best, a graph of $f$ in polar coordinates is used to suggest that these maxima go to zero as $r\to 0$.

Since $\sin \left(u\right)\doteq u-u^3\/3\!$, and $\cos \left(u\right)\doteq 1-u^2\/2\!\+u^4\/4\!$, the approximating rational function is

$\frac{x\left(x y-x^3{y}^3\/6\right)}{2-\left(1-x^2\/2\right)-\left(1-y^2\/2\right)}\=-\frac{1}{3}\frac{x^{2}y\left(x^2{y}^2-6\right)}{x^2\+y^2}$ 

for which the bivariate limit at the origin is

Since the bivariate limit of $f$ at the origin is 0, the required extension is

$g\left(x\,y\right)\=\{\begin{array}{cc}f\left(x\,y\right)& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

Indeed, $\mathrm{limit}\left(f\left(x\,y\right)\,\left\{x\=0\,y\=0\right\}\right)$ = $0$.