Let $f: \mathrm{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $\mathrm{f}(3)=18, \mathrm{f}^{\prime}(3)=0$ and $\mathrm{f}^{\prime \prime}(3)=4$.\\ Then $\lim _{x \rightarrow 1}\left(\log _{e}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)$ is equal to :