$F(x,y) =x^2 y /x^2+y^2 , (0,0) $. The answer is that F is not continues at $(0,0)$ because

  1. $\lim (x,y)>(0,0) ~ f(x,y)$ does not exist
  2. $(0,0) \in D_f$

where $D_f$ is the domain of $f$.

I try a method where you let $y=g_1(x)=x$ And $y=g_2(x)=x^2$ which is it to curves that pass the point $(0,0)$ etc And the limit is equal which means that the function is exist !

And why does point $(0,0)$ belong to $D_f$ ?

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  • $\begingroup$ Wow the English part and the math part are both not that readable. Would you please take some more time to rephrase the whole question so that your chance of getting a proper help will be much higher? $\endgroup$ – Megadeth Mar 17 '17 at 17:58
  • 1
    $\begingroup$ Am sorry am not English but I tried my best:) $\endgroup$ – Shouq Here Mar 17 '17 at 18:29
  • $\begingroup$ You may want to use \frac{}{} to avoid confusion. $\endgroup$ – user242756 Mar 17 '17 at 18:38

I do not have enough reputation to leave a comment. But, I think something is wrong with the way you wrote that function.

Your function simplifies to $F(x,y) = y + y^2$. This is continuous at $(0,0)$.

  • $\begingroup$ How it's simplifies as y+y^2 ? $\endgroup$ – Shouq Here Mar 17 '17 at 18:18
  • $\begingroup$ $F(x,y)=x^2y/x^2+y^2$. The first term is $x^2y/x^2 = y$. So $F(x,y)=y+y^2$ $\endgroup$ – JahKnows Mar 17 '17 at 18:23
  • $\begingroup$ How is x square equal y $\endgroup$ – Shouq Here Mar 17 '17 at 18:28
  • $\begingroup$ $\frac{x^2y}{x^2} = \frac{x^2}{x^2} \times y = 1 \times y = y$. $\endgroup$ – JahKnows Mar 17 '17 at 18:32

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