How to determine limits and continuity at for $f(x,y) = \frac{x^2 y}{x^4 + y^2}$for (x,y) ≠ (0,0) and f(0,0)=0 [closed]

I've just started to learn about multivariate function and limits and I'm not sure how to start of this question off. I'm supposed to -

determine the limit of f(x,y) as x,y approaches (0,0) along the line y=mx for $$f(x,y) = \frac{x^2 y}{x^4 + y^2}$$ for (x,y) ≠ (0,0) and f(0,0)=0

determine whether it is continuous at (0,0).

Any tips or ideas on how to get started and figure this out would be appreciated

Solution 1:

Consider, the path $y=mx^{2}$ and approach towards the origin along $y=mx^2$.

Then, \begin{align}f(x, mx^2)&= \frac{x^2 (mx^2)}{x^4 + (mx^2)^2}\\ &=\frac{mx^4}{x^4(1+m^2)}\\ &=\frac{m}{1 + m^2} \end{align}

This implies the limit at $(0, 0) $ is path dependent, choosing different values of $m$ we get different limit.

But if a limit of a function of two variables exists at a point, then limit along all path exists and all are equal.

For continuity, choose a sequence $(\frac{1}{n}, \frac{1}{n^2}) $

Then, $ (\frac{1}{n}, \frac{1}{n^2})\to (0, 0) $

But, $f(\frac{1}{n}, \frac{1}{n^2})=\frac{1}{2} \neq f(0, 0) $

Hence, $f$ is not continuous at $(0, 0) $