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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 $0$.

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  • $\begingroup$ Your question is a special case of a more general question which the OP answered, & asked to confirm if correct, in When does the limit $\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}$ exist?. In your case, $k = 6, l = 1, p = 4, q = 2$. As the OP showed, there's no limit if $\frac{k}{p}+\frac{l}{q}\le 2$. In your case, the left side is $\frac{6}{4} + \frac{1}{2} = 2$, so there's no limit. $\endgroup$ – John Omielan Jun 11 at 4:35
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No, this is not correct. Your first equality does not hold, nor do your later inequalities (remember that if $a < b$ then $\frac1a > \frac1b$).

To explore this further, consider what happens if you approach the origin:

a) along the curve $y = x^2$

b) along the curve $y = 0$

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actually the above limit does not exist because if we approach the origin (0,0) along the curve y =mx^2 (upward parabola) the limit depends upon m so taking different values of m we get different limit enter image description here

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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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