To show that $L$ is the bivariate limit at the origin, find $\mathrm{\&delta;}\left(\mathrm{\&epsilon;}\right)$ so that $\sqrt{x^2\&plus;y^2}<\mathrm{\&delta;}\Rightarrow \left|f\left(x\&comma;y\right)-L\right|<\mathrm{\&epsilon;}$.

Consider, then, the following annotated estimate for $\left|f\left(x\&comma;y\right)-0\right|$ = $\left|f\left(x\&comma;y\right)\right|$.

Consequently, $\mathrm{\&delta;}\&equals;\mathrm{\&epsilon;}\&sol;2$, that is, $\sqrt{x^2\&plus;y^2}<\mathrm{\&epsilon;}\&sol;2\Rightarrow \left|f\left(x\&comma;y\right)\right|<\mathrm{\&epsilon;}$.

Figure 3.2.18(a) compares $2\sqrt{x^2\&plus;y^2}$ with $\left|f\left(x\&comma;y\right)\right|$, the first in green, the second, in red. The green surface  lies above the red surface, indicating that near the origin, $2\sqrt{x^2\&plus;y^2}$ is greater than $\left|f\left(x\&comma;y\right)\right|$.

Maple corroborates these results by computing the bivariate limit, here accessed through the Context Panel.