(14) Given an ellipse centered at the origin $O$, with foci on the $x$-axis, and an eccentricity of $\frac{\sqrt{3}}{2}$, the ellipse passes through the point $\left(\sqrt{2}, \frac{\sqrt{2}}{2}\right)$. Let a line $l$ that does not pass through the origin $O$ intersect the ellipse at points $P$ and $Q$, and the slopes of the lines $OP$, $PQ$, and $OQ$ form a geometric sequence. Find the range of the area of $\triangle OPQ$.
(14) According to the problem, the equation of the ellipse is set as
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0),
$$

From $\left\{\begin{array}{l}\frac{c}{a}=\frac{\sqrt{3}}{2}, \\ \frac{2}{a^{2}}+\frac{1}{2 b^{2}}=1,\end{array}\right.$ we get $\left\{\begin{array}{l}a=2 \text {, } \\ b=1 .\end{array}\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$.
According to the problem, the slope of the line $l$ exists and is not 0, so we can set the equation of the line $l$ as $y=k x+m(m \neq 0), P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right)$ satisfy
$$
\left\{\begin{array}{l}
y=k x+m \\
x^{2}+4 y^{2}-4=0,
\end{array}\right.
$$

Eliminating $y$, we get $\left(1+4 k^{2}\right) x^{2}+8 k m x+4\left(m^{2}-1\right)=0$.
$$
\Delta=64 k^{2} m^{2}-16\left(1+4 k^{2}\right)\left(m^{2}-1\right)=16\left(4 k^{2}-m^{2}+1\right)>0 \text {, }
$$

and $x_{1}+x_{2}=\frac{-8 k m}{1+4 k^{2}}, x_{1} x_{2}=\frac{4\left(m^{2}-1\right)}{1+4 k^{2}}$.
$$
y_{1} y_{2}=\left(k x_{1}+m\right)\left(k x_{2}+m\right)=k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2} \text {. }
$$

Since the slopes of the lines $O P 、 P Q 、 O Q$ form a geometric sequence, we have
$$
\frac{y_{1}}{x_{1}} \cdot \frac{y_{2}}{x_{2}}=\frac{k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}}{x_{1} x_{2}}=k^{2},
$$

which means $\frac{-8 k^{2} m^{2}}{1+4 k^{2}}+m^{2}=0$, and since $m \neq 0$, we have $k^{2}=\frac{1}{4}$, i.e., $k= \pm \frac{1}{2}$.
Since the slope of the line $O Q$ exists and is not 0, and $\Delta>0$, we get $0<m^{2}<2$ and $m^{2} \neq 1$.
Let $d$ be the distance from point $O$ to the line $l$, then
$$
\begin{aligned}
S_{\triangle O P Q} & =\frac{1}{2} d|P Q|=\frac{1}{2} \frac{|m|}{\sqrt{1+k^{2}}} \sqrt{1+k^{2}}\left|x_{1}-x_{2}\right| \\
& =\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{m^{2}\left(2-m^{2}\right)},
\end{aligned}
$$

Therefore, the range of $S_{\triangle O P Q}$ is $(0,1)$.