It is known that when finding the limits of the multivariable function we have to find the limits using different paths .

Take the path along the x-axis ,then \(\displaystyle{\left({x},{y}\right)}\to{\left({x},{0}\right)}\)

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={0}\)

Take the path along the y-axis ,then

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={1}\)

Using the different paths , the values of the limits is different , then the limit does not exist.

Thus, the given function \(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}\) has no limits

Take the path along the x-axis ,then \(\displaystyle{\left({x},{y}\right)}\to{\left({x},{0}\right)}\)

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={0}\)

Take the path along the y-axis ,then

\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={1}\)

Using the different paths , the values of the limits is different , then the limit does not exist.

Thus, the given function \(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}\) has no limits