91影视

We have given expression:$$\begin{gathered} f(x,y)=\left\{\frac{\sin \left(x^2+y^2\right)}{x^2+y^2},if\left(x.y\right)鈮�\left(0,0\right)\right\} \\ \left\{1,if\left(x.y\right)=\left(0,0\right)\right\} \end{gathered}$$

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

The domain of a sine 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.

That is $x^2+y^2鈮�0$

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

Domain$\left(f\right)=f^{2}or\left\{\left(x,y\right):x,y鈭�f\right\}$

Substitute the value of $x=r\cos 胃$androle="math" localid="1653491811677" $y=r\sin 胃$

$$\begin{gathered} \underset{(x,y)鈫�\left(0,0\right)}{\lim }\frac{\sin \left(x^2+y^2\right)}{x^2+y^2}=\underset{r鈫�0}{\lim }\frac{Sin{(r\cos 胃)}^2+{(r\sin 胃)}^2}{{\left(r\cos 胃\right)}^2+{\left(r\sin 胃\right)}^2} \\ =\underset{r鈫�0}{\lim }\frac{Sin(r^2\left({\cos }^2胃+{\sin }^2胃\right))}{r^2\left(\cos {胃}^2+\sin {胃}^2\right)} \\ =\underset{r鈫�0}{\lim }\frac{Sin{r}^2\left(1\right)}{r^2\left(1\right)} \\ =\underset{r鈫�0}{\lim }\frac{Sin{r}^2}{r^2} \\ =1 \end{gathered}$$

Hence, there is no point of discontinuity for the given number.