Answer
(a)$$S_{\triangle PAO}=\frac{x}{2}$$ $$S_{\triangle PBO}=\frac{x^2}{2}$$
(b)
$$\begin{array}{|c|c|c|c|c|c|} \hline
x & 4 & 2 & 1 & 0.1 & 0.01 \\ \hline
\text{Area } \triangle PAO & 2 & 1 & 0.5 & 0.05 & 0.005 \\ \hline
\text{Area } \triangle PBO & 8 & 2 & 0.5 & 0.005 & 0.00005 \\ \hline
a(x) & 4 & 2 & 1 & 0.1 & 0.01 \\ \hline
\end{array}$$
$$\lim_{x \to 0^+}a(x)=0$$
Work Step by Step
(a)
The height of the triangle $\triangle PAO$ (drawn to the base $OA$) equals the $x$-component of the point $P(x,x^2)$, so we can obtain the area of the tringle $\triangle PAO$ as follows.$$OA=1, \quad h=x \quad \Rightarrow \quad S_{\triangle PAO}=\frac{h \cdot OA}{2}=\frac{x}{2}$$Similarly, the height of the triangle $\triangle PBO$ (drawn to the base $OB$) equals the $y$-component of the point $P(x,x^2)$, so we can obtain the area of the tringle $\triangle PBO$ as follows.$$OB=1, \quad h=x^2 \quad \Rightarrow \quad S_{\triangle PBO}=\frac{h \cdot OB}{2}=\frac{x^2}{2}$$
(b)
$a(x)$ equals $\frac{S_{\triangle PBO}}{S_{\triangle PAO}}$. So, by part (a) we obtain$$\lim_{x \to 0^+}a(x)= \lim_{x \to 0^+}\frac{\frac{x^2}{2}}{\frac{x}{2}}=\lim_{x \to 0^+}x=0.$$
