Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$ $$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$
I need to find the limit at (0,0) point
I applied the exponent rule and got $$e^{x^2y^2ln(x^2+y^2)}$$
and now with chain rule, I need to find the limit of $${x^2y^2ln(x^2+y^2)}$$
and how? :D
There isn't L'Hôpital's rule for multivariable function, right?

We have that
$$ {(x^2+y^2)}^{x^2y^2}=e^{x^2y^2 \log(x^2+y^2)} \to 1$$
indeed
$$x^2y^2 \log(x^2+y^2)=(x^2+y^2) \log(x^2+y^2) \frac{x^2y^2}{x^2+y^2} \to 0\cdot 0=0 $$
since by $t=x^2+y^2 \to 0$ by standard limits
$$(x^2+y^2) \log(x^2+y^2) =t\log t \to 0$$
and since $x^2+y^2 \ge 2xy$
$$0\le\frac{x^2y^2}{x^2+y^2} \le \frac{x^2y^2}{2xy} =\frac12 xy \to 0$$
