Calculating limit $\lim_{x,y\to \{0;0\} }(x^2+y^2)^{x^2 y^2} $ I have tried this:
$x=\rho \cos(\phi)$,
$y=\rho \sin(\phi)$. So we have:
$$\lim_{\rho \to 0}(\rho^2)^{\rho^4\cos^2(\phi)\sin^2(\phi)}=\left\langle{0^0}\right\rangle=\exp(\lim_{x \to 0}(x\cos^2(\phi)\sin^2(\phi)2\ln(x))=\exp(0)=1$$
but Wolfram says that its 0, am i wrong?

Recall that in one variable, $\lim_{u\to 0^+} u^u =1.$ In your problem the expression can be written
$$[(x^2+y^2)^{x^2+y^2}]^{x^2y^2/(x^2+y^2)}.$$
The term inside the brackets $\to 1$ by the above. So now just show that the outside exponent $\to 0$ to show the desired limit equals $1^0 = 1.$
