Answer
a. See the table, $-1.000$
b. See the graph, $-1.0$
c. $-1$
Work Step by Step
a. From the table, $\lim\limits_{x \to -2}\frac{x^2+3x+2}{2-|x|}\approx -1.000$
b. From the graph, $\lim\limits_{x \to -2}\frac{x^2+3x+2}{2-|x|}\approx -1.0$
c. $\lim\limits_{x \to -2}\frac{x^2+3x+2}{2-|x|}=\lim\limits_{x \to -2}\frac{(x+1)(x+2)}{2+x}=\lim\limits_{x \to -2}(x+1)=-1$
