91Ó°ÊÓ

We have given expression:$$\begin{gathered} f(x,y)=\left[\frac{xy}{x^2+y^2},if(x,y)≠(0,0)\right] \\ \left[0,if\left(x,y\right)=(0,0)\right] \end{gathered}$$

The given function is a piece wise function, with a break point at $(x,y)=(0,0)$

The domain of a polynomial function is the entire set of real numbers.

Hence, it never constraints the domain of the function.

Also the rational expression means that the denominator can not be equal to 0.

$x^2+y^2≠0$

The left side of this inequality is sum of two squares.

Hence, it is always positive.

The only case where it is not so, when$x=y=0$but that is not the case with the given function.

Hence, the function is defined for all the real values.

Thus, the domain of the function is given by the set is$\left\{\left(x,y\right):x,y∈f\right\}$.

Substitute the value of $x=r\cos \mathrm{θ}$and $y=r\sin \mathrm{θ}$

$$\begin{gathered} \underset{(x,y)→\left(0,0\right)}{\lim }\frac{xy}{x^2+y^2}=\underset{r→0}{\lim }\frac{\left(r\cos \mathrm{θ}\right)\left(r\sin \mathrm{θ}\right)}{{\left(r\cos \mathrm{θ}\right)}^2+{\left(r\sin \mathrm{θ}\right)}^2} \\ =\underset{r→0}{\lim }\frac{r^2\cos \mathrm{θ}\sin \mathrm{θ}}{r^2\cos {\mathrm{θ}}^2+r^2\sin {\mathrm{θ}}^2} \\ =\underset{r→0}{\lim }\frac{r^2\cos \mathrm{θ}\sin \mathrm{θ}}{r^2\left(\cos {\mathrm{θ}}^2+\sin {\mathrm{θ}}^2\right)} \\ =\underset{r→0}{\lim }\frac{r^2\cos \mathrm{θ}\sin \mathrm{θ}}{r^2\left(1\right)} \\ =\cos \mathrm{θ}\sin \mathrm{θ} \end{gathered}$$

Thus, the given function is continuous over$f^2-(0,0)or\left\{\left(x,y\right):\left(x,y\right)≠\left(0,0\right)\right\}$