<s>[INST] Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit? 
$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
I'm having trouble with proving that this limit really goes to $0$
thank you
 [/INST] $$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
Method 1: Using the two path test
Along the path of $y=0$, we have
$$\lim\limits_{(x,0)\to (0,0)} \frac{x^3 + 0^3}{x^2 + 0}=\lim\limits_{x\to 0} \frac{x^3}{x^2}=\lim\limits_{x\to 0} x=0$$
Along the path of $y=p(x)-x^2$, where $p(x)$ is a polynomial that passes through the origin, we have
$$\lim\limits_{(x,p(x)-x^2)\to (0,0)} \frac{x^3 + (p(x)-x^2)^3}{x^2 + p(x)-x^2}$$
$$=\lim\limits_{x\to 0} \frac{x^3 -x^6+ 3x^4p(x)-3x^2p(x)^2+p(x)^3}{p(x)}$$
$$=\lim\limits_{x\to 0} \left[\frac{x^3 -x^6}{p(x)}+ 3x^4-3x^2p(x)+p(x)^2\right]$$
Let $p(x)=x^3-x^6$, then
$$\lim\limits_{x\to 0} \left[1+ 3x^4-3x^2(x^3-x^6)+(x^3-x^6)^2\right]=1$$
Since we have different values along different paths, we can conclude that
$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$
Method 2: Using polar coordinates
$$\lim\limits_{r\to 0^+} \frac{r^3\cos^3\phi + r^3\sin^3\phi}{r^2\cos^2\phi + r\sin\phi}$$
$$=\lim\limits_{r\to 0^+} r^2\left(\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right)$$
Now lets attempt to find bounds that are independent of $\phi$. Since 
$$\left|\cos^3\phi + \sin^3\phi\right|\leq 1$$
We have
$$r^2\left|\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right|\leq \frac{r^2}{\left|r\cos^2\phi + \sin\phi\right|}$$
Note that $\phi$ is a variable and we cannot treat it as a constant. Since the right hand side is dependent on $\phi$, then we can conclude that
$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$
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