## Calculus (3rd Edition)

$$c= -4$$
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 $$c= -4$$