TITLE: Multivariable limit $\lim_ {{(x,y)} \to {(0,0)}} \frac{xy^2\ln\frac{|x|}{|y|}}{{(x^2+y^2)}^{\frac 12}}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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

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