I want to see that the limit $$\lim_{(x,y)\to\ (0,0)}xy\log(x^2+y^2)=0$$ So my idea is to bound: $$0\leq|xy\log(x^2+y^2)|$$ by something that tends to zero when $x,y\to(0,0)$, but I'm stuck with: $$0\leq|xy\log(x^2+y^2)|\leq|(x^2+y^2)\log(x^2+y^2)|$$ Since I don't know what to do with $\log(x^2+y^2)$.
Note: Working with real variables.
Write your term in the form $$\frac{xy}{x^2+y^2}\times(x^2+y^2)\ln(x^2+y^2)$$ Now use that $$\lim_{x\to 0^+ }x\ln(x)=...$$=
Let's transform $f(x,y) = xy\log(x^2+y^2)$ into a more useful form: Let $$ x = r \cos \theta\\ y = r\sin \theta \\ f(x,y) = r^2\cos\theta\sin\theta \log(r^2) = r^2\cos (2\theta)\log r $$
$\cos(2\theta)$ is bounded in absolute value being $\leq 1$ so $$ |f(x,y)| = |f(r,\theta)| \leq r^2\log r $$ This holds for all $x$ and $y$. Therefore, if $$ \lim_{r\to 0} r^2\log r = 0 $$ then $$ \lim_{x,y\to 0} f(x,y) = 0 $$ But writing $r^2 \log r = \frac{\log r}{r^{-2}}$ and appying L'Hopital's rule we see that limit is zero.
You can say when x and y, both goes to zero or very near to 0 then X is approximately equal to y. So replace y with x then take limit as x tends to 0
We have that
$$xy\log(x^2+y^2)=\frac{xy}{x^2+y^2}\cdot (x^2+y^2)\log(x^2+y^2) \to 0$$
indeed we can se for example by polar coordinates that
is bounded and since $t=x^2+y^2 \to 0$
