Answer
$$c= -4$$
Work Step by Step
Given $$
\lim _{x \rightarrow 1} \frac{x^{2}+3 x+c}{x-1}
$$
Since the limit exist when $ x-1$ is a factor of $ x^{2}+3 x+c$ , consider
\begin{align*}
x^{2}+3 x+c&= (x-1) (x+a)\\
&= x^2+ (a-1)x-a
\end{align*}
by comparing, we get $$a-1=3\ \ \ \to \ \ a= 4$$
and $$ c= -4$$
