# Is this proof of the $\lim_{(x,y)\to (0,0))} \frac{x^6y}{x^8+y^4}$ correct? [duplicate]

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 $$0$$.

• 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. – John Omielan Jun 11 at 4:35

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 $$y = 0$$

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

NO. $$(|x|^6|y| and $$|x|^8+|y|^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 $$\frac {x^6y}{x^8+y^4}=0.$$