Multivariable limit $\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}$

If we examine what happens when we approach $(0,0)$ along the line $x=0$ (or $y=0$), we get: $$ \lim_{(x,y)\to(0,0)} (x²+y²)^{x²y²}= \lim_{y\to0} (0+y^2)^{0}= 1 $$

meaning that if the limit exists it must be $1$.

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?

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

$$\lim_{r\to0}r\ln r=0$$