## University Calculus: Early Transcendentals (3rd Edition)

(a) $\lim_{x\to-2^+}(x+3)\frac{|x+2|}{x+2}=1$ (b) $\lim_{x\to-2^-}(x+3)\frac{|x+2|}{x+2}=-1$
- For $x\gt-2$: $$x+2\gt0\Rightarrow|x+2|=x+2$$ - For $x\lt-2$: $$x+2\lt0\Rightarrow|x+2|=-(x+2)$$ a) $$A=\lim_{x\to-2^+}(x+3)\frac{|x+2|}{x+2}$$ As $x\to-2^+$, we only consider the values of $x$ to the right of $-2$, which means $x\gt-2$, so $|x+2|=x+2$. Therefore, $$A=\lim_{x\to-2^+}(x+3)\frac{x+2}{x+2}$$ $$A=\lim_{x\to-2^+}(x+3)$$ $$A=-2+3=1$$ b) $$B=\lim_{x\to-2^-}(x+3)\frac{|x+2|}{x+2}$$ As $x\to-2^-$, we only consider the values of $x$ to the left of $-2$, which means $x\lt-2$, so $|x+2|=-(x+2)$. Therefore, $$B=\lim_{x\to-2^-}(x+3)\frac{-(x+2)}{x+2}$$ $$B=\lim_{x\to-2^-}-(x+3)$$ $$B=-(-2+3)=-1$$