For each $t \in (0,1)$, the surface $P_t \in \mathbb{R}^3$ is defined by $P_t = \{(x,y,z) : (x^2 +y^2)z = 1, \, t^2\le x^2 +y^2 \le 1\}$. Let $A_t$ be the surface area of $P_t$. Then $\lim_{\{t\rightarrow 0^{+}\}}A_t$ exists or not.
After evaluation, I found that $A_t=$ 
.
But how to determine the existence of the limits
Got it. After converting to polar coordinates, $A_t=\int\int_{t^2\le r^2\le1}\sqrt{1+\frac{4}{r^6}}$. Which can be seen as the volume of a solid bounded above by the surface $z=\sqrt{1+\frac{4}{r^6}}$ and below by the region $t^2\le r^2\le 1$ in the $xy$-plane. But as $t$ tends to 0, the height of the solid is unbounded. Hence, the volume is unbounded. The limit does not exist.