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Let $L = \lim_{x\rightarrow1}\ln(\frac{f(x+2)}{f(3)})^{\frac{18}{(x-1)^{2}}}$.
Then $L = \lim_{x\rightarrow1} \frac{18}{(x-1)^2} \ln(\frac{f(x+2)}{f(3)})$.
Let $x-1 = h$, so $x = h+1$. As $x \rightarrow 1$, $h \rightarrow 0$.
Then $L = \lim_{h\rightarrow0} \frac{18}{h^2} \ln(\frac{f(h+3)}{f(3)})$.
Using Taylor series expansion for $f(h+3)$ around $h=0$, we have $f(h+3) = f(3) + hf'(3) + \frac{h^2}{2}f''(3) + O(h^3)$.
Given $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$, we have $f(h+3) = 18 + 0 + \frac{h^2}{2}(4) + O(h^3) = 18 + 2h^2 + O(h^3)$.
So, $L = \lim_{h\rightarrow0} \frac{18}{h^2} \ln(\frac{18 + 2h^2 + O(h^3)}{18}) = \lim_{h\rightarrow0} \frac{18}{h^2} \ln(1 + \frac{2h^2}{18} + O(h^3))$.
Using the approximation $\ln(1+x) \approx x$ for small $x$, we have $L = \lim_{h\rightarrow0} \frac{18}{h^2} (\frac{2h^2}{18} + O(h^3))$.
Therefore, $L = \lim_{h\rightarrow0} \frac{18}{h^2} \cdot \frac{2h^2}{18} = 2$.
Correct Answer: 2
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