Chapter 2: Limits and Continuity - Section 2.4 - One-Sided Limits - Exercises 2.4 - Page 75: 17

a. $1$ b. $-1$

a. For $x\to-2^+$, we have $x+2\gt0$, thus $\lim_{x\to-2^+}(x+3)\frac{|x+2|}{x+2}=\lim_{x\to-2^+}(x+3)\frac{x+2}{x+2}=\lim_{x\to-2^+}(x+3)=-2+3=1$ b. For $x\to-2^-$, we have $x+2\lt0$, thus $\lim_{x\to-2^-}(x+3)\frac{|x+2|}{x+2}=\lim_{x\to-2^+}(x+3)\frac{-(x+2)}{x+2}=-\lim_{x\to-2^+}(x+3)=-(-2+3)=-1$