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 Quote by Galileo You've shown that: |x+2|<1 <=> -5
Alright this kinda clarifies my question, so -5 is like the min? Which would grant -5<x-2<5--since epsilon can never be E<0 - the absolute value is necessary? Sorry if I come off a bit slow, but it's just that I'm trying to justify every step so I completely understand the concept.