Answer
$c=-1$ and $c=6$
Work Step by Step
$$\lim _{x \rightarrow c} \frac{x^{2}-5 x-6}{x-c}$$
This limit will exist, provided that $x-c$ is a factor of the numerator. (Otherwise there will be an infinite discontinuity at $x=c$.) Since $$x^{2}-5 x-6=(x+1)(x-6),$$ then we know that this occurs for $c=-1$ and $c=6$.
