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How can I demonstrate the continuity of this function. If $a>0$, \begin{align} f(x,y) = \left\{ \begin{array}{lll} \dfrac{x^3y}{ax^4+y^4} & \text{if} & (x,y) \ne (0,0) \\ 0 & \text{if} & (x,y) = (0,0) \end{array} \right. \end{align}

I suppose I'm gonna have to calculate the limit of the function when $(x,y) \rightarrow (0,0)$. But I don't understand what that's gonna bring to my proof.

How can I continue my proof?

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Hint :

Approach to the point $(0,0)$ along the path $y=mx$.

Which shows that $f$ is discontinuous at $(0,0)$.

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  • $\begingroup$ Showing discontinuity at (0,0)............. $\endgroup$ – DanielWainfleet Oct 20 '15 at 17:27
  • $\begingroup$ Can you show how to calculate this limit? I suppose $$lim \frac{x^3(mx)}{ax^4+(mx)^4}=\frac{mx^4}{ax^4+m^4x^4}=\frac{m}{a+m}$$. $\endgroup$ – hlapointe Oct 20 '15 at 17:31
  • $\begingroup$ That limit depends on $m$ ..and so $f$ is discontinuous at $(0,0)$ $\endgroup$ – Empty Oct 20 '15 at 17:32
  • $\begingroup$ Do you know the basic definition of continuity & discontinuity of function of several variable ? $\endgroup$ – Empty Oct 20 '15 at 17:33
  • $\begingroup$ Not really. I supposed that is the same way of a function of one variable? $\endgroup$ – hlapointe Oct 20 '15 at 17:34

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