Let f: R →(0, ∞) be a twice differentiable function such that f(3)=18, f^′(3)=0 and f^′ ′(3)=4. Then lim_x → 1( log_e…

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