e) $lim_{x\rightarrow - 4}\frac{|x + 4|}{2x+8}$
e) $lim_{x\rightarrow - 4}\frac{|x + 4|}{2x+8}$
Answer
Explanation:
Step1: Factor the denominator
Factor (2x + 8) as (2(x + 4)). So the limit becomes (\lim_{x\rightarrow - 4}\frac{|x + 4|}{2(x + 4)}).
Step2: Consider left - hand and right - hand limits
Left - hand limit ((x\rightarrow - 4^{-})):
When (x\rightarrow - 4^{-}), (x+4<0), so (|x + 4|=-(x + 4)). Then (\lim_{x\rightarrow - 4^{-}}\frac{|x + 4|}{2(x + 4)}=\lim_{x\rightarrow - 4^{-}}\frac{-(x + 4)}{2(x + 4)}=-\frac{1}{2}).
Right - hand limit ((x\rightarrow - 4^{+})):
When (x\rightarrow - 4^{+}), (x + 4>0), so (|x + 4|=x + 4). Then (\lim_{x\rightarrow - 4^{+}}\frac{|x + 4|}{2(x + 4)}=\lim_{x\rightarrow - 4^{+}}\frac{x + 4}{2(x + 4)}=\frac{1}{2}). Since the left - hand limit (-\frac{1}{2}\neq\frac{1}{2}) (the right - hand limit), the limit (\lim_{x\rightarrow - 4}\frac{|x + 4|}{2(x + 4)}) does not exist.
Answer:
The limit does not exist.