Does $\lim_{(x,y) \to (0,0)} xy^4 / (x^2 + y^8)$ exist?

Question

From this question on answers.yahoo, the guy says the following limit does not exist: $$\lim_{(x,y) \to (0,0)} \frac{xy^4}{x^2 + y^8},$$ then on wolfram, it says the limit is equal to $0$. When I did it myself, I tried approaching $(0,0)$ from the $x$-axis, $y$-axis, $y=x$, $y=x^2$. They all equal $0$.

But when I tried the squeeze theorem, I got $y^8 \leq x^2 + y^8$, therefore $0 \leq |xy^4/(x^2+y^8)| \leq |\dfrac{x}{y^4}|$, and the latter does not exist for $(x, y) \to (0,0)$.

So does the original limit exist or not? I'm getting contradicting information from various sources. Also, if it doesn't exist (it looks like it doesn't... I think), how would I go about proving that it doesn't?

One example more (already close to 10 just in the last 4-5 months that I can remember) that shows that WA cannot be trusted in many cases... — DonAntonio, Nov 13 '13 at 11:50
Possible duplicate of Prove that $\lim_{(x,y)\rightarrow(0,0)} \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}}$ does not exist — Arnaud D., Dec 03 '19 at 18:11

score 5 · Answer 1 · answered Nov 13 '13 at 10:05

5

Approaching $(0,0)$ along the curve $x=y^4$ you get

$$ \frac{xy^4}{x^2+y^8} = \frac{y^8}{y^8 + y^8} = \frac12 $$

which clearly does not tend to $0$ as $y \to 0$. Since you get different values along different paths, the limit doesn't exist.

answered Nov 13 '13 at 10:05

mrf

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I think is $x=y^4$ – Nov 13 '13 at 10:08
@nanchangjian Yes, thank you. – mrf Nov 13 '13 at 10:08

score 1 · Answer 2 · answered Nov 13 '13 at 10:07

1

Hint: let $x=ky^4(k\neq 0)$,so $$\lim_{(x,y)\to(0,0)}\dfrac{xy^4}{x^2+y^8}=\dfrac{k}{k^2+1}$$

answered Nov 13 '13 at 10:07

Lucian · Answer 3 · 2013-11-13T10:11:04.077

-1

$a=x,b=y^4=>\lim_{(0,0)}\frac{ab}{a^2+b^2}=\lim_{(0,0)}\frac{a/b}{(a/b)^2+1}=$ Indeterminate, since for instance we could have various ratios between x and $y^4$ as they both approach $0$.

edited Nov 13 '13 at 10:11

answered Nov 13 '13 at 10:03

Lucian

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Why $\lim_0\frac{a/b}{(\frac ab)^2+1}=\lim_0\frac ab$? – hhsaffar Nov 13 '13 at 10:05
We simplify both sides through $b^2$. – Lucian Nov 13 '13 at 10:08
Oh, sorry, I thought you were asking something else. It's not, I've corrected it. The point is that a/b can be anything. – Lucian Nov 13 '13 at 10:16

Does $\lim_{(x,y) \to (0,0)} xy^4 / (x^2 + y^8)$ exist?

3 Answers3