For $g$ to be differentiable at the origin, $\mathrm{\Δ} g\equiv g\left(0\+h\,0\+k\right)-g\left(0\,0\right)\=g\left(h\,k\right)$ must assume the form

$g_x\left(0\,0\right) h\+g_y\left(0\,0\right) k\+\mathrm{\η}$$\left(h\,k\right)\cdot \sqrt{h^2\+k^2}$ 

where $\mathrm{\η}\to 0$ as $\left(h\,k\right)\to \left(0\,0\right)$. Since $g_x\left(0\,0\right)\=g_y\left(0\,0\right)\=0$ from Example 4.11.1, it follows that

where $\mathrm{\λ}\left(x\,y\right)\=\sqrt{h^2\+k^2}\sin \left(\frac{1}{h^2\+k^2}\right)$. Since $\mathrm{\λ}$ is the product of a bounded factor and a factor that goes to zero, $\mathrm{\λ}\to 0$ as $\left(h\,k\right)\to \left(0\,0\right)$. Hence, setting $\mathrm{\η}\=\mathrm{\λ}$ implies that $g$ is differentiable at the origin.$$