# Precise definition of a limit

by hmm?
Tags: definition, limit, precise
 P: 19 Hello, I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks. Prove that lim x->-2 (x^2-1)=3 0<|f(x)-L|
HW Helper
P: 2,002
 Quote by hmm? Assume |x+2|<1 so -1
You've shown that: |x+2|<1 <=> -5<x-2<-3
This implies that |x-2|<5, simply because x-2<-3 means that surely x-2<5, so that -5<x-2<5, or |x-2|<5.

You wrote |x-2|<-3, which is never true becuase the left side is positive and the right side isn't. Did you mean |x-2|<3?
HW Helper
PF Gold
P: 4,771
Small detail that is important:

 Prove that lim x->-2 (x^2-1)=3 0<|f(x)-L|
You only want to prove that "|f(x)-L|<epsilon whenever 0<|x-a|<delta".

In other words, f(x) can be = L.

 P: 4 Precise definition of a limit Perhaps this way would work: Prove that lim x->-2 (x^2-1)=3 let E > 0 be given and: |f(x) - L| = |x^2 - 4| = |(x - 2)(x + 2)| = |x - 2||x + 2| if 0 < |x + 2| < 1 then, -3 < x < -1 |x - 2| < 5 < 5|x + 2| therefore: if 0 < |x + 2| < 1 and 5|x + 2| < E then by transitivity of <(less than), |f(x) - L| < E or reworded: if D = minimum(1, E/5), then |f(x) - L| < E (where E = epsilon; D = delta)
 P: 4 Sorry this part: if 0 < |x + 2| < 1 then, -3 < x < -1 |x - 2| < 5 is a "side part"
P: 19
 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.

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