Consider the function whose formula is f(x) = rac16−x4x2−4. Find f(2) and _x -2 f(x).

Substitute $x=2$ into the function:

$$f(2) = \frac{16-(2)^4}{(2)^2-4} = \frac{16-16}{4-4} = \frac{0}{0}$$

Since this is an indeterminate form, the function is undefined at $x=2$.

$$f(2) \text{ is undefined}$$

First, attempt to substitute $x=-2$ into the function:

$$f(-2) = \frac{16-(-2)^4}{(-2)^2-4} = \frac{16-16}{4-4} = \frac{0}{0}$$

This is an indeterminate form, so we need to simplify the function by factoring.

The numerator is a difference of squares: $16-x^4 = (4-x^2)(4+x^2)$.
The denominator is also a difference of squares: $x^2-4$.
So, we can write $f(x)$ as:

$$f(x) = \frac{(4-x^2)(4+x^2)}{x^2-4}$$

Notice that $(4-x^2) = -(x^2-4)$. Substitute this into the expression:

$$f(x) = \frac{-(x^2-4)(4+x^2)}{x^2-4}$$

For $x^2-4 \neq 0$ (i.e., $x \neq 2$ and $x \neq -2$), we can cancel the common factor $(x^2-4)$:

$$f(x) = -(4+x^2)$$

Now, find the limit as $x \to -2$ for the simplified function:

$$\lim_{x \to -2} f(x) = \lim_{x \to -2} -(4+x^2)$$

Substitute $x=-2$ into the simplified expression:

$$= -(4+(-2)^2)$$ $$= -(4+4)$$ $$= -8$$

The final answers are:
$f(2)$ is undefined
$\lim_{x \to -2} f(x) = \boxed{-8}$