I have to say if the following limit exist or not. $$\lim_{(x,y)\to(0,0)}\frac{3xy+y^2}{x^3-y}$$ So long I have graph the function and use Wolfram's tool and everything points that the value is zero. I have tried to use $\epsilon- \delta$ to proof, but no success. I tried using polar coordinates and get: $$\lim_{r\to 0^+}\frac{3r^2 \cos(x)\sin(x)+r^2\sin(x)^2}{r^3\cos(x)^3-r\sin(x)}= \lim_{r\to0^+} \frac{3r \cos(x)\sin(x)+r\sin(x)^2}{r^2\cos(x)^3-\sin(x)} $$ And then put the quotient apart: $$\lim_{r\to0^+}(3r \cos(x)\sin(x)+r\sin(x)^2)=0$$ Independent of the angle $x$ $$\lim_{r\to0^+}(r^2\cos(x)^3-\sin(x))=-\sin(x)$$ which gives zero as a result, except in the case $\sin(x)=0$ But that is the case $y=0$ And it is easy to see that in that case the value is also zero.
My problem is the following, I'm not very good at proofs, so the hole thing can be horribly wrong. Is this a correct way of proving things? Is there a better way to say that the limit is zero? Does it matter that it is not define in $y=x^3$ ? And if you are generous enough I also have a problem with: $$\lim_{(x,y)\to(0,0)}\frac{x^2-xy}{x+y}$$ But they are so similar that a solution of the first will give me an idea of how to make the second one.
Any help will be great. Thank you for your time and help. ((:
Note that on the path $y=x^3-x^5$, we have
$$\begin{align} \frac{3xy+y^2}{x^3-y}&=\frac{3x^4-2x^6-2x^8+x^{10}}{x^5}\\\\ &=3x^{-1}-2x-2x^3+x^5 \tag 1 \end{align}$$
The limit of the right-hand side of $(1)$ fails to exist as $x\to 0$ since the limits from the right and left sides are unequal.
Therefore, the limit of interest fails to exist.