<s>[INST] Under these conditions, determine the polynomial in two variables. Let $f(x,y)=(x^2+y^2+2)^{xy}$ defined on $\mathbb R^2$ and let $g(x,y)$ be a degree $2$ polynomial in two variables.
Suppose $\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{g(x,y)-f(x,y)}{x^2+y^2}=0$. Then, determine $g(x,y)$.

Here is what I did so far.
Let $g(x,y)=a_0 x^2+a_1 y^2 +a_3 xy + a_4 x+ a_5 y+ a_6.$
From supposition, $\lim_{(x,y)\to (0,0)}g(x,y)-f(x,y)=0$ has to hold. And
since $\lim g(x,y)=a_6, \lim f(x,y)=1$, I get $a_6=1.$
If I use polar coordinates,
\begin{align}
&\dfrac{g(x,y)-f(x,y)}{x^2+y^2}\\
&=\dfrac
{r^2(a_0 \cos^2 \theta + a_1 \sin^2 \theta +a_3 \cos\theta \sin \theta)+ r(a_4 \cos \theta + a_5 \sin \theta)+1-(r^2+2)^{r^2 \cos \theta \sin \theta}}
{r^2}\\ 
&=a_0 \cos^2 \theta + a_1 \sin^2 \theta +a_3 \cos\theta \sin \theta+\dfrac{1}{r}(a_4 \cos \theta + a_5 \sin \theta)+ \dfrac{1-(r^2+2)^{r^2 \cos \theta \sin \theta}}{r^2}
\end{align}
When $r\to 0$ with $\theta=0,$ this has to converge to $0$ so
$a_0+\dfrac{a_4}{r}+\dfrac{1-1}{r^2}\to 0$ as $r\to 0$.
Thus $a_4=a_0=0.$
Similarly, when $r\to 0$ with $\theta=\frac{\pi}{2},$ this has to converge to $0$ so
$a_1+\dfrac{a_5}{r}+\dfrac{1-1}{r^2}\to 0$ as $r\to 0$.
Thus $a_5=a_1=0$.
(I'm not sure the argument above is correct.)
Thus, all that is left is to determine $a_3.$
Now, $g(x,y)=a_3 xy +1$. I'm stacked here.
How can I determine $a_3$ ?
 [/INST] Recall that for $t\to 0$, $e^t=1+t+O(t^2)$ and $\ln(1+t)=t+o(t)$.
Therefore,for $(x,y)\to (0,0)$,
\begin{align}(x^2+y^2+2)^{xy}&=\exp(xy\ln(x^2+y^2+2))\\
&=1+xy\ln(x^2+y^2+2)+O(x^2y^2)\\
&=1+xy\left(\ln(2)+\ln\left(1+\frac{x^2+y^2}{2}\right)\right)+o(x^2+y^2)\\
&=1+xy\left(\ln(2)+\frac{x^2+y^2}{2}+o(x^2+y^2)\right)+o(x^2+y^2)\\
&=1+\ln(2)xy+o(x^2+y^2).\end{align}
Hence $g(x,y)=1+\ln(2)xy$. In particular $a_3=\ln(2)$.
</s>