Puisque \(\left|\frac{x^{2}-y^{2}}{x^{2} + y^{2}}\right|\) reste borné, \[\mathrm{lim}_{(x,y)\to (0,0)}f(x,y)= \mathrm{lim}_{(x,y)\to (0,0)}xy\frac{x^{2}-y^{2}}{x^{2} + y^{2}} = \mathrm{lim}_{(x,y)\to (0,0)}xy =0\] d’où \(f\) est continue en \((0,0)\). De même, \[\begin{align*} \mathrm{lim}_{(x,y)\to (0,0)}\frac{f(x,y)}x&= \mathrm{lim}_{(x,y)\to (0,0)}y\frac{x^{2}-y^{2}}{x^{2} + y^{2}} = \mathrm{lim}_{(x,y)\to (0,0)}y =0 \\ \mathrm{lim}_{(x,y)\to (0,0)}\frac{f(x,y)}y&= \mathrm{lim}_{(x,y)\to (0,0)}x\frac{x^{2}-y^{2}}{x^{2} + y^{2}} = \mathrm{lim}_{(x,y)\to (0,0)}x =0 \end{align*}\] d’où les dérivées partielles \(\frac{\partial f}{\partial x}(0,0)\) et \(\frac{\partial f}{\partial y}(0,0)\) existent, et \(\frac{\partial f}{\partial x}(0,0)=0\) et \(\frac{\partial f}{\partial y}(0,0)=0\). En plus, en dehors de l’origine, \[\begin{align*} \frac{\partial f}{\partial x}&= \frac{f(x,y)}x + xy \frac{\partial}{\partial x} \frac{x^{2}-y^{2}}{x^{2} + y^{2}} = \frac{f(x,y)}x + 4\frac{x^{2}y^{3}}{(x^{2} + y^{2})^2} \\ \frac{\partial f}{\partial y}&= \frac{f(x,y)}y + xy \frac{\partial}{\partial y} \frac{x^{2}-y^{2}}{x^{2} + y^{2}} = \frac{f(x,y)}x + 4\frac{x^{3}y^{2}}{(x^{2} + y^{2})^2} . \end{align*}\] Puisque \[\mathrm{lim}_{(x,y) \to (0,0)}\frac{x^{2}y^{3}}{(x^{2} + y^{2})^2}=0,\ \mathrm{lim}_{(x,y) \to (0,0)}\frac{x^{3}y^{2}}{(x^{2} + y^{2})^2}=0,\] il s’ensuit que \[\mathrm{lim}_{(u,v) \to (0,0)}\frac{\partial f}{\partial x}(u,v) =0,\ \mathrm{lim}_{(u,v) \to (0,0)}\frac{\partial f}{\partial y}(u,v) =0,\] d’où les dérivées partielles \(\frac{\partial f}{\partial x}\) et \(\frac{\partial f}{\partial y}\) sont continues en \((0,0)\).

exo7 1800