Classify the discontinuity of z(x) at x = -1

Factor the numerator: $x^2 - 1 = (x-1)(x+1)$, so for $x \neq -1$, $z(x) = \frac{(x-1)(x+1)}{x+1} = x - 1.$ This simplification is valid everywhere except at $x = -1$, where the original expression has a $0/0$ form.

Since $z(x) = x - 1$ for all $x$ near $-1$ (but not equal to it), $\lim_{x \to -1} z(x) = -1 - 1 = -2$. This limit exists and is finite, but $z(-1)$ itself is undefined since it produces $0/0$. Because the limit exists but $f(a)$ is not defined, condition 1 fails, and the discontinuity is removable — the graph of $z$ has a hole at $(-1, -2)$.

The hole can be patched by defining the function's value at negative one to equal negative two. This matches the earlier finding that the limit as x approaches negative one is negative two, even though z was originally undefined there. Assigning this value makes the function continuous at that point, since all three continuity conditions would then be satisfied.