If f(3) = 18, f′(3) = 0 and f″(3) = 4. Then the value of

lim_x→1 [ f(x + 2) / f(3) ]^{18/(x² − 1)}

The given question asks for the limit:

$L\; =\; lim\_\{x\; \to \; 1\}\; [\; (f(x\; +\; 2))\; /\; (f(3))\; ]^(18\; /\; (x\; -\; 1)^2)$

Given the values:
$f(3)\; =\; 18$
$f\text{'}(3)\; =\; 0$
$f\text{'}\text{'}(3)\; =\; 4$

Step 1: Identify the form of the limit
As $x\; \to \; 1$, the base term behaves as follows:
$lim\_\{x\; \to \; 1\}\; [\; (f(x\; +\; 2))\; /\; (f(3))\; ]\; =\; (f(3))\; /\; (f(3))\; =\; 18\; /\; 18\; =\; 1$
The exponent term tends to $\infty$ as $x\; \to \; 1$.
Therefore, the limit is of the indeterminate form $1^\infty$.

Step 2: Simplify the limit using the exponential property
For a limit of the form $lim\_\{x\; \to \; a\}\; [g(x)]^(h(x))$ where $g(x)\; \to \; 1$ and $h(x)\; \to \; \infty$, the limit is given by:
$e^(K)$
where:
$K\; =\; lim\_\{x\; \to \; 1\}\; [\; g(x)\; -\; 1\; ]\; ·\; h(x)$

Substituting $g(x)\; =\; (f(x\; +\; 2))\; /\; (f(3))$ and $h(x)\; =\; 18\; /\; (x\; -\; 1)^2$:
$K\; =\; lim\_\{x\; \to \; 1\}\; [\; (f(x\; +\; 2))\; /\; (f(3))\; -\; 1\; ]\; ·\; [\; 18\; /\; (x\; -\; 1)^2\; ]$

Since $f(3)\; =\; 18$:
$K\; =\; lim\_\{x\; \to \; 1\}\; [\; (f(x\; +\; 2)\; -\; 18)\; /\; 18\; ]\; ·\; [\; 18\; /\; (x\; -\; 1)^2\; ]$
$K\; =\; lim\_\{x\; \to \; 1\}\; (f(x\; +\; 2)\; -\; f(3))\; /\; (x\; -\; 1)^2$

Step 3: Evaluate the limit using L'Hôpital's Rule
Since $f(3)\; =\; 18$ and $f\text{'}(3)\; =\; 0$, the limit $K$ is of the indeterminate form $0/0$.
Differentiating the numerator and the denominator with respect to $x$ (L'Hôpital's Rule):
$K\; =\; lim\_\{x\; \to \; 1\}\; (f\text{'}(x\; +\; 2))\; /\; (2(x\; -\; 1))$

Since $f\text{'}(3)\; =\; 0$, this is still of the form $0/0$. Applying L'Hôpital's Rule a second time:
$K\; =\; lim\_\{x\; \to \; 1\}\; (f\text{'}\text{'}(x\; +\; 2))\; /\; 2$
$K\; =\; (f\text{'}\text{'}(3))\; /\; 2$

Substitute the given value $f\text{'}\text{'}(3)\; =\; 4$:
$K\; =\; 4\; /\; 2\; =\; 2$

Thus, the value of the limit's exponent parameter is 2.