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?
 A: 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 $1^0 = 1.$
A: You can indeed compute
$$
\lim_{(x,y)\to(0,0)}x^2y^2\log(x^2+y^2)=
\lim_{\rho\to0}2\sin^2\phi\cos^2\phi\rho^4\log\rho
$$
Now consider
$$
0\le 2\sin^2\phi\cos^2\phi\rho^4\log\rho\le 2\rho^4\log\rho
$$
and that
$$
\lim_{\rho\to0}\rho\log\rho=0
$$
By the squeeze theorem, the limit is $0$.
Thus your original limit is $\exp(0)=1$.
A: If there is a limit it must be $1$, since if we set $y=0$, we get $x^0=1$ (for $x\ne 0$).  The issue is whether there is a limit.  Your calculations seem reasonable, noting that $(\rho)^{X}=\exp(X\log\rho)$, where $X=2\rho^4 \sin^2(\phi)\cos^2(\phi)$. We have $|\rho\log\rho|\to0$ as $\rho\to0+$, so a fortiori $|X\log\rho|\to 0$, and so $\exp(X\log \rho)\to 1$.  I don't know, however, what $\langle{0^0}\rangle$ means.
