# By considering different path of approach, show that the function f(x,y)=(x^2+y)/y has no limis (x,y) ->(0,0)

Multivariable functions
By considering different path of approach, show that the function $$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}$$ has no limis $$\displaystyle{\left({x},{y}\right)}\to{\left({0},{0}\right)}$$

2021-01-03
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