Step-by-Step Calculation
1. Simplifying the Expression:
Let $L = \lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$
Using log property $\log(a^b) = b \log(a)$:
$$L = \lim_{x \to 1} \frac{18}{(x-1)^2} \log_e \left( \frac{f(2+x)}{f(3)} \right)$$
2. Checking the Form:
As $x \to 1$, the expression becomes $\frac{18}{0} \log_e(\frac{f(3)}{f(3)}) = \infty \times \log_e(1) = \infty \times 0$.
This is a $\frac{0}{0}$ indeterminate form if written as:
$$L = 18 \lim_{x \to 1} \frac{\log_e f(2+x) – \log_e f(3)}{(x-1)^2}$$
3. Applying L’Hôpital’s Rule (First Time):
Differentiating numerator and denominator w.r.t $x$:
$$L = 18 \lim_{x \to 1} \frac{\frac{1}{f(2+x)} \cdot f'(2+x)}{2(x-1)}$$
$$L = 9 \lim_{x \to 1} \frac{f'(2+x)}{f(2+x)(x-1)}$$
4. Applying L’Hôpital’s Rule (Second Time):
Since $f'(3) = 0$, it is still $\frac{0}{0}$ form.
$$L = 9 \lim_{x \to 1} \frac{f”(2+x)}{f'(2+x)(x-1) + f(2+x)(1)}$$
5. Substituting the Values:
Now put $x = 1$:
$$L = 9 \left[ \frac{f”(3)}{f'(3)(0) + f(3)(1)} \right]$$
$$L = 9 \left[ \frac{4}{0 + 18} \right] = 9 \times \frac{4}{18} = 2$$
Theoretical Concepts & Calculations
Double Differentiability and Limits
A function is twice differentiable at a point if its first derivative is also differentiable at that point. In this problem, the existence of $f”(3)$ allows us to apply L’Hôpital’s Rule twice. This is a common pattern in competitive exams where a limit involves a second-order change. The property $\lim_{x \to a} \frac{g(x)}{h(x)}$ can be solved by $\lim_{x \to a} \frac{g”(x)}{h”(x)}$ provided the first derivatives also result in $0/0$.
Logarithmic Expansion Method
Alternatively, one could use Taylor series expansion for $f(2+x)$ around $x=1$. Let $2+x = 3+h$ where $h = x-1$.
$f(3+h) = f(3) + hf'(3) + \frac{h^2}{2}f”(3) + \dots$
Substituting $f(3)=18, f'(3)=0, f”(3)=4$:
$f(3+h) = 18 + 0 + \frac{h^2}{2}(4) = 18 + 2h^2$.
The expression becomes $\log( \frac{18+2h^2}{18} ) = \log(1 + \frac{h^2}{9})$.
Using $\log(1+u) \approx u$ for small $u$:
$L = \lim_{h \to 0} \frac{18}{h^2} \cdot \frac{h^2}{9} = 2$.
Properties of Twice Differentiable Functions
Twice differentiable functions ensure that the graph is smooth and its curvature is well-defined. Here, $f'(3)=0$ and $f”(3)=4 > 0$ indicates that the function has a local minimum at $x=3$. This physical interpretation helps in visualizing why the function stays positive ($(0, \infty)$) near the point of evaluation.