Evaluating the Limit Expression
The problem asks for the evaluation of the limit: $L = \lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$
Simplifying the Limit
Using the property of logarithms $\log(a^b) = b \log(a)$, we can rewrite the expression: $L = \lim_{x \to 1} \frac{18}{(x-1)^2} \log_e \left( \frac{f(2+x)}{f(3)} \right)$
To simplify further, let $y = 2+x$. As $x \to 1$, the variable $y$ approaches $3$. The term $x-1$ can be written as $y-3$. Substituting these into the limit expression: $L = \lim_{y \to 3} \frac{18}{(y-3)^2} \log_e \left( \frac{f(y)}{f(3)} \right)$
Using the logarithm property $\log(a/b) = \log(a) - \log(b)$: $L = \lim_{y \to 3} \frac{18 \left( \log_e f(y) - \log_e f(3) \right)}{(y-3)^2}$
Applying L'Hôpital's Rule
As $y \to 3$, the limit takes the indeterminate form $\frac{0}{0}$ (since $f(3)=18$, $\log_e f(3)$ is finite, and the numerator approaches $18(\log_e f(3) - \log_e f(3)) = 0$, while the denominator approaches $(3-3)^2 = 0$). We can apply L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to $y$.
- Numerator derivative: $ \frac{d}{dy} \left[ 18 (\log_e f(y) - \log_e f(3)) \right] = 18 \left( \frac{f'(y)}{f(y)} \right) $
- Denominator derivative: $ \frac{d}{dy} \left[ (y-3)^2 \right] = 2(y-3) $
Applying L'Hôpital's Rule: $L = \lim_{y \to 3} \frac{18 f'(y) / f(y)}{2(y-3)} = \lim_{y \to 3} \frac{9 f'(y)}{f(y)(y-3)}$
The limit is still in the indeterminate form $\frac{0}{0}$ because $f'(3)=0$ and $f(3)=18$. We apply L'Hôpital's Rule a second time.
- Numerator derivative: $ \frac{d}{dy} [9 f'(y)] = 9 f''(y) $
- Denominator derivative: $ \frac{d}{dy} [f(y)(y-3)] = f'(y)(y-3) + f(y)(1) $ (using the product rule)
Applying L'Hôpital's Rule again: $L = \lim_{y \to 3} \frac{9 f''(y)}{f'(y)(y-3) + f(y)}$
Substituting Given Function Values
Now, we substitute $y=3$ into the expression and use the provided values: $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. $L = \frac{9 f''(3)}{f'(3)(3-3) + f(3)}$ $L = \frac{9 \cdot 4}{0 \cdot (0) + 18}$ $L = \frac{36}{0 + 18}$ $L = \frac{36}{18}$ $L = 2$
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