Question

${\lim }_{x\to -1}\frac{x^2+3x+2}{x^2-4x+3}$

• A. 0
• B. 1/2
• C. 2
• D. 3

Answer 1

Answer: (B)

${\lim }_{x\to -1}\frac{x^2+3x+2}{x^2+4x+3}={\lim }_{x\to -1}\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+3\right)}$
= ${\lim }_{x\to -1}\frac{x+2}{x+3}=\frac{-1+2}{-1+3}=\frac{1}{2}$