the upper indices in the limit denote right - and left - hand limits. if both the right - and left - hand…

the upper indices in the limit denote right - and left - hand limits. if both the right - and left - hand limits are not equal, then the limit does not exist. a) lim_{x→ - 18^+} \frac{|x + 18|}{x + 18} b) lim_{x→ - 18^-} \frac{|x + 18|}{x + 18} c) lim_{x→ - 18} \frac{|x + 18|}{x + 18}

the upper indices in the limit denote right - and left - hand limits. if both the right - and left - hand limits are not equal, then the limit does not exist. a) lim_{x→ - 18^+} \frac{|x + 18|}{x + 18} b) lim_{x→ - 18^-} \frac{|x + 18|}{x + 18} c) lim_{x→ - 18} \frac{|x + 18|}{x + 18}

Answer

Explanation:

Step1: Analyze the right - hand limit (a)

When $x\to - 18^{+}$, $x + 18>0$, so $|x + 18|=x + 18$. Then $\lim_{x\to - 18^{+}}\frac{|x + 18|}{x + 18}=\lim_{x\to - 18^{+}}\frac{x + 18}{x + 18}=1$.

Step2: Analyze the left - hand limit (b)

When $x\to - 18^{-}$, $x + 18<0$, so $|x + 18|=-(x + 18)$. Then $\lim_{x\to - 18^{-}}\frac{|x + 18|}{x + 18}=\lim_{x\to - 18^{-}}\frac{-(x + 18)}{x + 18}=-1$.

Step3: Analyze the two - sided limit (c)

Since the right - hand limit $\lim_{x\to - 18^{+}}\frac{|x + 18|}{x + 18}=1$ and the left - hand limit $\lim_{x\to - 18^{-}}\frac{|x + 18|}{x + 18}=-1$, and $1\neq - 1$, the two - sided limit $\lim_{x\to - 18}\frac{|x + 18|}{x + 18}$ does not exist.

Answer:

a) 1 b) - 1 c) Does not exist