1. The problem asks to find the limit as $x$ approaches $-1$ of the function $$f(x) = \frac{x^2 - 1}{x + 1}.$$\n\n2. First, note that directly substituting $x = -1$ into $f(x)$ gives $$f(-1) = \frac{(-1)^2 - 1}{-1 + 1} = \frac{1 - 1}{0} = \frac{0}{0},$$ which is undefined. This is an indeterminate form, so we need to simplify the expression.\n\n3. Factor the numerator: $$x^2 - 1 = (x - 1)(x + 1).$$\n\n4. Substitute back into $f(x)$: $$f(x) = \frac{(x - 1)(x + 1)}{x + 1}.$$\n\n5. We can cancel the common factor $x + 1$ in numerator and denominator, but remember this is only valid for $x \neq -1$ because division by zero is undefined. Using the cancellation notation: $$f(x) = \frac{(x - 1)\cancel{(x + 1)}}{\cancel{(x + 1)}} = x - 1, \quad x \neq -1.$$\n\n6. Now, the limit as $x$ approaches $-1$ of $f(x)$ is the same as the limit of $x - 1$ as $x$ approaches $-1$: $$\lim_{x \to -1} f(x) = \lim_{x \to -1} (x - 1) = -1 - 1 = -2.$$\n\n7. The value $-1$ is the point $x$ approaches, not the value of the function at $x = -1$. The function is undefined at $x = -1$ because the original denominator is zero there. However, the limit exists and equals $-2$ because the simplified function $x - 1$ approaches $-2$ as $x$ approaches $-1$.\n\n8. In summary, the limit is about the behavior of $f(x)$ near $x = -1$, not the function's value at $x = -1$. That's why the limit is $-2$ and not zero.