Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 69: 7

$4$

$x^{4} - 1 = (x - 1)(x^3 + x^2 + x + 1)$ so: $$\frac{x^{4}-1}{x-1} = x^3 + x^2 + x + 1.$$ Now we have: $$\begin{aligned} \lim\limits_{x \to 1^{+}}\frac{x^{4}-1}{x-1}& = \lim\limits_{x \to 1^{+}}(x^3 + x^2 + x + 1)\\ & = 1^3 + 1^2 + 1 + 1 \\ &= 4. \end{aligned}$$