Determine the domains of the functions in Exercises 47–56, and find where the functions are continuous.

$$\begin{gathered} f(x,y)=\left\{e^{-1/x^2+y^2}if(x.y)\ne (0.0)\right\} \\ \left\{0if\left(x,y\right)=\left(0,0\right)\right\} \end{gathered}$$

We have given expression: $$\begin{gathered} f(x,y)=\left\{e^{-1/x^2+y^2}if(x.y)\ne (0.0)\right\} \\ \left\{0if\left(x,y\right)=\left(0,0\right)\right\} \end{gathered}$$

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

The given function can be write as

An exponential function as well as its reciprocal is defined for all the values at which the exponent is defined.

The exponent itself is a rational expression.

The rational expression means that the denominator cannot be equal to 0.

That is $x^2+y^2\ne 0$

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 as,

$f^{2}or\left\{\left(x+y\right):x.y\in f\right\}$

Substitute the value of $x=r\cos \theta$and $y=r\sin \theta$

$$\begin{gathered} \underset{\left(x,y\right)\to \left(0,0\right)}{\lim }\frac{1}{e^{1/x^2+y^2}}=\underset{r\to 0}{\lim }\frac{1}{e^{{\displaystyle \frac{1}{{(r\cos \theta )}^2+{(r\sin \theta )}^2}}}} \\ =\underset{r\to 0}{\lim }\frac{1}{e^{{\displaystyle \frac{1}{{r^2(\cos \theta }^2+{\sin \theta )}^2}}}} \\ =\underset{r\to 0}{\lim }\frac{1}{e^{{\displaystyle \frac{1}{r^2}}}} \\ =0 \end{gathered}$$

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