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

Question
Multivariable functions
asked 2021-01-02
By considering different path of approach, show that the function \(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}\) has no limis \(\displaystyle{\left({x},{y}\right)}\to{\left({0},{0}\right)}\)

Answers (1)

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
0

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