<s>[INST] Is this proof of the $\lim_{(x,y)\to (0,0))} \frac{x^6y}{x^8+y^4}$ correct? Let $\varepsilon>0$ be arbitrary. Note that $||(x,y)-(0,0)||<\delta \implies \sqrt{x^2+y^2}<\delta$, which in turn yields $|x|<\delta$ and $|y|<\delta$. 
Now, let $\delta=\varepsilon$ and assume that $||(x,y)-(0,0)||<\delta$. From what we just proved, $|x|<\varepsilon$ and $|y|<\varepsilon$. Finally, we have:
$$\bigg|\frac{x^6y}{x^8+y^4}\bigg|=\frac{|x|^6|y|}{|x|^8+|y|^4}<\frac{\varepsilon^7}{\varepsilon^8+\varepsilon^4}<\frac{\varepsilon^7}{\varepsilon^4}<\varepsilon^3$$
Since $\varepsilon^3\to 0$ as $\varepsilon\to0$, the limit is $0$.
 [/INST] NO. $(|x|^6|y|<e^7$ and $|x|^8+|y|^4<e^8+e^4)$ does NOT imply that $\frac {|x|^6|y|}{|x|^8+|y|^4}<\frac {e^7}{e^8+e^4}$ for the same reason that $(2<3$ and $1<8)$ does not imply that $\frac {2}{1}<\frac {3}{8}.$
There is no limit. If $0\ne y=x^2$ then $\frac {x^6y}{x^8+y^4}=\frac {1}{2}$ but if $x=0\ne y$ then $\frac {x^6y}{x^8+y^4}=0.$
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