Evaluating a limit by factoring to remove a common factor

Substituting $x = -1$ directly gives $\frac{-1+1+1-1}{1-1} = \frac{0}{0}$, an indeterminate form, so we cannot evaluate the limit by direct substitution and must simplify the expression algebraically first. Factor the numerator by grouping: $x^3 + x^2 - x - 1 = x^2(x+1) - (x+1) = (x+1)(x^2 - 1)$. Notice that $x^2 - 1$ also factors as $(x-1)(x+1)$, so the numerator becomes $(x+1)^2(x-1)$.

The denominator factors as $x^2 - 1 = (x-1)(x+1)$. Writing the whole expression as $\frac{(x+1)^2(x-1)}{(x-1)(x+1)},$ we can cancel the common factors $(x-1)$ and one factor of $(x+1)$, valid for $x \neq -1, 1$ near $x=-1$, leaving $x+1$. Substituting $x = -1$ into this simplified expression gives $-1 + 1 = 0$.