# 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: 2,004
 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?
PF Patron
HW Helper
P: 4,755
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.

 Related Discussions Precalculus Mathematics Homework 22 Calculus & Beyond Homework 6 Calculus & Beyond Homework 3 Introductory Physics Homework 2 Calculus 1