## Calculus (3rd Edition)

$c=3$
Since $\frac{1}{x-1}-\frac{c}{x^{3}-1}=\frac{x^{2}+x+1-c}{(x-1)\left(x^{2}+x+1\right)}$ In order for the limit to exist as $x \rightarrow 1$, the numerator must evaluate to 0 at $x=1 .$ Thus, we must have $3-c=0,$ which implies that $c=3$.