$g(x) \cdot g(y)=g(x)+g(y)+g(x y)-2$
Put $x=1, y=2$, then
$$
\begin{array}{l}
g(1) \cdot g(2)=g(1)+g(2)+g(2)-2 \\
5 g(1)=g(1)+5+5-2 \\
4 g(1)=8 \quad \therefore g(1)=2
\end{array}
$$
Put $\mathrm{y}=\frac{1}{\mathrm{x}}$ in equation (1), we get
$$
\begin{array}{l}
\mathrm{g}(\mathrm{x}) \cdot \mathrm{g}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{g}(\mathrm{x})+\mathrm{g}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{g}(1)-2 \\
\mathrm{~g}(\mathrm{x}) \cdot \mathrm{g}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{g}(\mathrm{x})+\mathrm{g}\left(\frac{1}{\mathrm{x}}\right)+2-2 \\
{[\because \mathrm{g}(1)=2]}
\end{array}
$$
This is valid only for the polynomial
$$
\therefore \quad \mathrm{g}(\mathrm{x})=1 \pm \mathrm{x}^{\mathrm{n}}
.....(2)$$
$\operatorname{Now} g(2)=5$
(Given)
$\begin{aligned} \therefore 1 & \pm 2^{n}=5 &\text { [Using equation (2)] }\\ & \pm 2^{\mathrm{n}}=4, \Rightarrow 2^{n}=4,-4 \end{aligned}$
Since the value of $2^{n}$ cannot be-Ve.
So, $2^{n}=4, \Rightarrow n=2$
Now, put $\mathrm{n}=2$ in equation (2), we get $g(x)=1 \pm x^{2}$
$$
\begin{aligned}
\therefore \quad \operatorname{Lt}_{x \rightarrow 3} g(x) &=\operatorname{Lt}_{x \rightarrow 3}\left(1 \pm x^{2}\right)=1 \pm(3)^{2} \\
&=1 \pm 9=10,-8
\end{aligned}
$$