Evaluating a limit by factoring to remove a 0/0 form

Direct substitution at $x=1$ gives $\frac{1-1}{1-1} = \frac{0}{0}$, an indeterminate form, so we cannot evaluate the limit yet by direct substitution and must simplify the expression first. Factor the numerator using the difference of squares: $x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$.

The expression becomes $\frac{(x-1)(x+1)(x^2+1)}{x-1},$ and since $x \neq 1$ near the limit, we can cancel the common factor $(x-1)$, leaving $(x+1)(x^2+1)$, which is a polynomial and hence continuous everywhere. By the limit laws (product and sum rules), we can now substitute $x = 1$ directly: $(1+1)(1^2+1) = 2 \cdot 2 = 4$.

The answer is polynomial. After canceling the common factor of x minus one, the remaining expression is the product of x plus one and x squared plus one, which is a polynomial. This is important because polynomials are continuous everywhere, so direct substitution can be applied immediately afterward.