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

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.

Hence,

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

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 $0$.

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

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