Answer
(a)$$L_{\triangle PAO}=\sqrt{x^4-x^2+1}+\sqrt{x^4+x^2}+1$$ $$L_{\triangle PBO}=\sqrt{x^4+x^2-2x+1}+\sqrt{x^4+x^2}+1$$
(b)
$$\begin{array}{|c|c|c|c|c|c|} \hline
x & 4 & 2 & 1 & 0.1 & 0.001 \\ \hline
\text{Perimeter } \triangle PAO & 33.016597 & 9.0776872 & 3.4142136 & 2.0955364 & 2.0099505 \\ \hline
\text{Perimeter } \triangle PBO & 33.771243 & 9.5952416 & 3.4142136 & 2.0005543 & 2.0000005 \\ \hline
r(x) & 0.97765419 & 0.94606135 & 1 & 1.0474779 & 1.004975 \\ \hline
\end{array}$$
$$\lim_{x \to 0^+}r(x)= 1$$
Work Step by Step
(a)
As we know, the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. So the perimeter of the triangle $\triangle PAO$ with the vertices $O(0,0)$, $A(0,1)$, and $P(x,x^2)$ can be obtained as follows.$$OA=\sqrt{(0-0)^2+(1-0)^2}=1, \quad OP=\sqrt{(x-0)^2+ (x^2-0)^2}=\sqrt{x^4+x^2}, \quad AP= \sqrt{(x-0)^2+(x^2-1)^2}=\sqrt{x^4-x^2+1} \quad \Rightarrow \quad L_{\triangle PAO}=AP+OP+OA=\sqrt{x^4-x^2+1}+\sqrt{x^4+x^2}+1.$$
Similarly, the perimeter of the triangle $\triangle PBO$ with the vertices $O(0,0)$, $B(1,0)$, and $P(x, x^2)$ can be obtained as follows.$$OB=\sqrt{(1-0)^2+(0-0)^2}=1, \quad OP=\sqrt{(x-0)^2+ (x^2-0)^2}=\sqrt{x^4+x^2}, \quad BP= \sqrt{(x-1)^2+(x^2-0)^2}=\sqrt{x^4+x^2-2x+1} \quad \Rightarrow \quad L_{\triangle PBO}=BP+OP+OB=\sqrt{x^4+x^2-2x+1}+\sqrt{x^4+x^2}+1.$$
(b)
$r(x)$ equals $\frac{L_{\triangle PAO}}{L_{\triangle PBO}}$. So, by part (a) we can obtain$$\lim_{x \to 0^+}r(x)=\lim_{x \to 0^+}\frac{\sqrt{x^4-x^2+1}+\sqrt{x^4+x^2}+1}{\sqrt{x^4+x^2-2x+1}+\sqrt{x^4+x^2}+1}=\frac{2}{2}=1.$$
