Multivariable limit $\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}$ I'm studying multivariable limits and I have a problem regarding this limit:
$$\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}.$$  I've found that it is equal to 1, by rewriting
the limit using $t = e^{\ln(t)}$. Despite this, the answer in the book is 0. What is the correct answer?
 A: Let $x=r\cos t$ and $y=r\sin t$ so
$$\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}=\lim_{r\to0}r^{\frac12r^4\sin^2(2t)}=1$$
because
$$\lim_{r\to0}r\ln r=0$$
A: The limit cannot be $0$ because if so, then
$$\sqrt{x^{2} + y^{2}} < \delta \Rightarrow |(x^{2} + y^{2})^{(xy)^{2}}| < \epsilon,$$
$$x^{2} + y^{2} < \delta^{2}; x^{2}, y^{2} < \delta^{2},$$
$$(x^{2} + y^{2})^{(xy)^{2}} < \delta^{2\delta^{4}} \leq \epsilon,$$
$$\delta \ln \delta = (\frac{\ln \epsilon}{2})^{1/4}.$$
If
$$\epsilon := 1/2$$
then
$(\ln \epsilon /2)^{1/4}$ becomes meaningless.
A: If we examine what happens when we approach $(0,0)$ along the line $x=0$ (or $y=0$), we get:
$$
\lim_{(x,y)\to(0,0)} \left(x^2+y^2\right)^{x^2y^2}=
\lim_{y\to0} \left(0+y^2\right)^{0}=
1
$$
meaning that if the limit exists it must be $1$.
A: Using polar coordinates, we get
$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies (x^2+y^2)^{x^2y^2}=r^{2r^4sin^2t\cos^2t}=r^{\frac12r^4\sin^22t}=e^{\frac12r^4\sin^22t\log r}\xrightarrow[r\to 0^+]{}e^0=1$$
so I'd go with you being correct and the book is wrong. What book is that, anyway?
