## Calculus, 10th Edition (Anton)

$x^{4}$ - 1 = (x - 1)(x³ + x² + x + 1), so: $\frac{x^{4}-1}{x-1}$ = x³ + x² + x + 1. Now we have: $\lim\limits_{x \to 1^{+}}\frac{x^{4}-1}{x-1}$ = $\lim\limits_{x \to 1^{+}}x³ + x² + x + 1$ = 1³ + 1² + 1 + 1 = 4