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Given is the following function $f(x,y)= \frac{x^3-xy^2}{\vert x \vert + y^2}$. How to proof that the limit $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist?

I have yet tried approaching zero in various directions, yet I have not been able to find two with different outcomes.

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    with $y=0$ set $x\to0^{\pm}$ gives two values. – 2017-02-21
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    @MyGlasses $0=0$ are the same value. ;-)) – 2017-02-21
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    Possibly test the limit as $x\to0$ and $y\to0$ seperate? Does saying $(x,y)\to (0,0)$ imply they are converging at the same rate? – 2017-02-21
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    @NickPavini It implies no such thing. But this limit does exist. – 2017-02-21
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    @Omry I was just brainstorming :) – 2017-02-21
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    @Dr.MV I saw it :) I was contemplating on doing this in a polar way, but that's easier than mine. – 2017-02-21

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The reason why you cannot solve it is the fact that the limit exists and is $0$:

$$0 \le \left| \frac {x^3 -xy^2} {|x| + y^2} \right| = \frac {|x^3 -xy^2|} {|x| + y^2} \le \frac {|x^3| + |xy^2|} {|x| + y^2} \le \frac {|x^3| + |xy^2|} {|x|} = |x^2| + |y^2| \to 0 ,$$

which shows (by the sandwich theorem) that

$$\left| \frac {x^3 -xy^2} {|x| + y^2} \right| \to 0 ,$$

whence

$$\frac {x^3 -xy^2} {|x| + y^2} \to 0 .$$

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The limit is $0$. To see this, we note that if $\sqrt{x^2+y^2}\le 1$, then $|x|+y^2\ge x^2+y^2$.

Hence, we have

$$\begin{align} \left|\frac{x^3-xy^2}{|x|+y^2}\right|&\le \frac{|x|\,|x^2-y^2|}{x^2+y^2}\\\\ &\le \frac{|x|(x^2+y^2)}{x^2+y^2}\\\\ &=|x| \end{align}$$

And we are done!

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    hmm how do we notice that $\sqrt{x^2+y^2}\le1$? – 2017-02-21
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    @NickPavini You are taking the limit in which they approach zero. – 2017-02-21
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    @NickPavini As $(x,y)\to (0,0)$, $\sqrt{x^2+y^2}\to 0$. – 2017-02-21
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    Oh ok. You factored a bit? And used some basic limit properties? – 2017-02-21
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    @NickPavini Which part of the solution do you not understand? – 2017-02-21
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    I dont understand how you isolated the square root, and I dont understand why you put everything in absolute value. I do not mean to be difficult so if I am too lost for this then by all means dont feel inclined to give me a whole lesson lol :) – 2017-02-21
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    @NickPavini If $|x|\le 1$, then $|x|\ge x^2$ and hence $|x|+y^2\ge x^2+y^2$. And the absolute values ensures that the term of interest doesn't "runaway to the left with negative values. – 2017-02-21
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    Ohh because they are converging at $0$? So $x$ and $y$ will be less than 1? I think I see it thank you:) – 2017-02-21
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    @NickPavini Well done! You have it now. – 2017-02-21