<s>[INST] Answer Says $\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6} = 0$. I say DNE. What did I do wrong? I was asked to find 
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6}$$ 
Observe that setting y=mx results in $$\lim_{(x,mx)\rightarrow(0,0)}\frac{x^3(mx)^2}{x^4+(mx)^6} = 0$$
The textbook solution then proved that the limit is 0 using the squeeze theorem.
However, I tried to set y=x^(4/6) and I got:
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3(x^{\frac{4}{6}})^2}{x^4+(x^{\frac{4}{6}})^6} = \lim_{x\rightarrow0}\frac{x^4}{x^4+x^4} = \frac{1}{2}$$
So I concluded that the limit does not exist. 
I am not convinced that my solution is correct, I would really appreciate to know the reason why I am wrong. 
Thank you
 [/INST] Another approach:
$$\frac{x^3y^2}{x^4+y^6}\stackrel{\text{Polar Coord.}}=\frac{r^5\cos^3\theta\sin^2\theta}{r^4(\cos^4\theta+r^2\sin^6\theta)}=r\frac{\cos^3\theta\sin^2\theta}{\cos^4\theta+r^2\sin^6\theta}\xrightarrow[r\to0]{}0$$
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