## Calculus (3rd Edition)

$f(-1)=\frac{(-1)^{2}+2(-1)+1}{-1+1}=\frac{0}{0}$ The function has the indeterminate form $\frac{0}{0}$ at x=-1. Transforming algebraically and canceling, we have $\frac{x^{2}+2x+1}{x+1}=\frac{(x+1)(x+1)}{x+1}=x+1$ Evaluating using continuity, we get $\lim\limits_{x \to -1}\frac{x^{2}+2x+1}{x+1}=\lim\limits_{x \to -1}(x+1)=-1+1=0$