I have to prove that the limit

$$\lim_ {{(x,y)} \to {(0,0)}} \frac{xy^2\ln\frac{|x|}{|y|}}{{(x^2+y^2)}^{\frac 12}}$$

does not exist. I've tried to find two different paths that show that the limit is divergent, but I couldn't find any. I've also tried to bound it, and it didn't work. Can somebody explain me how to do this? Thank you!

PS: This is part of a much larger excercise, I didn't include it because it was irrelevant. I just need to prove that it does not converges to zero.

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    $\begingroup$ I believe the limit is in fact $0$... $\endgroup$ – David H Aug 6 '14 at 22:43
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    $\begingroup$ Why do you think the limit does not exist? $\endgroup$ – Mhenni Benghorbal Aug 6 '14 at 22:48
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    $\begingroup$ Me too, have a you look at this i.stack.imgur.com/y7GOr.png It seems only the values for $(x,y) = (a,0)$ or $(0,a)$ do not exist, but the limits do. $\endgroup$ – flawr Aug 6 '14 at 22:48

The limit is $0$.

As it appears, the function is not defined for $x=0,y\neq0$ or $y=0,x\neq0$ but it could be extended continuously there since $z \ln|z| \rightarrow 0$ as $z\rightarrow 0$.

$$\lim_ {{(x,y)} \to {(0,0)}} \frac{xy^2\ln\frac{|x|}{|y|}}{{(x^2+y^2)}^{\frac 12}}=\lim_ {{(x,y)} \to {(0,0)}} \frac{xy\ln|x|-xy\ln |y|}{{[1+(x/y)^2]}^{\frac 12}}=0.$$

Note that

$$\lim_{x\rightarrow 0} x\ln|x|=0, \\ \lim_{y\rightarrow 0} y\ln|y|=0,\\ [1+(x/y)^2]^\frac1{2} \neq 0$$

Since, in this limit, $x$ and $y$ tend to $0$ together -- we cannot have $xy = 1$, for example.


$$\lim_{(x,y)\rightarrow (0,0)} xy\ln|x|-xy\ln |y|=0$$

So the numerator tends to $0$ regardless of the path and the denominator is never $0$. The denominator could approach $\infty$ as $y$ tends faster to $0$ than $x$, e.g., $y=x^2$. In that case , the limit of the ratio converges more rapidly to $0$.

  • $\begingroup$ @user84412: While reading RRL's answer here, I realized that my answer had a serious error and so have withdrawn it. The conclusion I was aiming towards was in fact wrong; I'd suggest you read this answer instead. $\endgroup$ – Semiclassical Aug 7 '14 at 14:31

First let me recall the result

$$ |\ln(x)|< \frac{1}{x},\quad x\sim 0. $$

$$\Bigg| \frac{xy^2\ln\frac{|x|}{|y|}}{{(x^2+y^2)}^{\frac 12}} \Bigg|. $$

I'll handle the case $|x|<|y|<1$ which gives

$$ \Big| \frac{x}{y} \Big|<1 \implies \ln\Big| \frac{x}{y} \Big| < \Big| \frac{y}{x} \Big|.$$

So we have

$$\Bigg| \frac{xy^2\ln\frac{|x|}{|y|}}{{(x^2+y^2)}^{\frac 12}} \Bigg| < \frac{|x|\,|y|^2\frac{|y|}{|x|}}{{(x^2+y^2)}^{\frac 12}} = \frac{|y|^3}{{(x^2+y^2)}^{\frac 12}} \leq (x^2+y^2) < \epsilon $$

$$ \implies \sqrt{x^2+y^2} < \sqrt{\epsilon} =\delta. $$

Note: The following inequalities are useful

$$ |y| = \sqrt{y^2} \leq \sqrt{x^2+y^2}. $$


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