We need to evaluate the limit of the function $\frac{x^2 - 4}{x^2 - 16}$ as $x$ approaches 4. That is, find $\lim_{x \to 4} \frac{x^2 - 4}{x^2 - 16}$.

1. Problem Description

We need to evaluate the limit of the function $\frac{x^2 - 4}{x^2 - 16}$ as $x$ approaches

4. That is, find $\lim_{x \to 4} \frac{x^2 - 4}{x^2 - 16}$.

2. Solution Steps

First, notice that if we directly substitute $x = 4$ into the expression, we get $\frac{4^2 - 4}{4^2 - 16} = \frac{16 - 4}{16 - 16} = \frac{12}{0}$, which is undefined. We can try to simplify the expression by factoring the numerator and denominator.

The numerator is a difference of squares:

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

The denominator is also a difference of squares:

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

So, the expression becomes:

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

Now we evaluate the limit:

$\lim_{x \to 4} \frac{(x - 2)(x + 2)}{(x - 4)(x + 4)}$.

When $x$ approaches 4, the numerator approaches $(4 - 2)(4 + 2) = (2)(6) = 12$, and the denominator approaches $(4 - 4)(4 + 4) = (0)(8) = 0$. Since we have a non-zero number divided by a number approaching 0, the limit will be either $\infty$ or $-\infty$. Let's investigate the sign of the expression as $x$ approaches 4 from the left and right.

If $x$ approaches 4 from the left ($x < 4$), then $x - 4$ is negative and $x + 4$ is positive. Also, $x - 2$ and $x + 2$ are positive. Therefore, the expression $\frac{(x - 2)(x + 2)}{(x - 4)(x + 4)}$ is negative.

If $x$ approaches 4 from the right ($x > 4$), then $x - 4$ is positive and $x + 4$ is positive. Also, $x - 2$ and $x + 2$ are positive. Therefore, the expression $\frac{(x - 2)(x + 2)}{(x - 4)(x + 4)}$ is positive.

Since the limit from the left is $-\infty$ and the limit from the right is $\infty$, the limit does not exist.

3. Final Answer

The limit does not exist.