We are asked to evaluate the limit: $\lim_{x \to 4} \frac{x-4}{x^2-16}$.

1. Problem Description

We are asked to evaluate the limit: $\lim_{x \to 4} \frac{x-4}{x^2-16}$.

2. Solution Steps

First, we can factor the denominator as a difference of squares:

$x^2 - 16 = (x-4)(x+4)$

Then, we rewrite the expression inside the limit:

$\frac{x-4}{x^2-16} = \frac{x-4}{(x-4)(x+4)}$

As $x$ approaches 4, but is not equal to 4, we can cancel out the $(x-4)$ terms:

$\frac{x-4}{(x-4)(x+4)} = \frac{1}{x+4}$

Now, we can evaluate the limit by substituting $x=4$:

$\lim_{x \to 4} \frac{1}{x+4} = \frac{1}{4+4} = \frac{1}{8}$

3. Final Answer

The limit is $\frac{1}{8}$.

So the answer is (D).