Answer
$\lim\limits_{x \to -1}\frac{2x^2+3x+1}{x^2+2x-3}=0.25$
Work Step by Step
$\lim\limits_{x \to -1}\frac{2x^2+3x+1}{x^2+2x-3}=\lim\limits_{x \to -1}\frac{(2x+1)(x+1)}{(x-3)(x+1)}=\lim\limits_{x \to -1}\frac{(2x+1)}{(x-3)}=\frac{\lim\limits_{x \to -1}{(2x+1)}}{\lim\limits_{x \to -1}{(x-3)}}=\frac{2\times(-1)+1}{-1-3}=\frac{-1}{-4}=0.25$
