# Precise definition of a limit

1. Oct 1, 2006

### hmm?

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|<epsilon whenever 0<|x-a|<delta

0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta
= |x^2-4|<epsilon
= |(x-2)(x+2)|<epsilon
= |x-2||x+2|<epsilon
= If |x-2|<C (C=constant) then |x-2||x+2|<C|x+2|
= C|x+2|<epsilon = |x+2|<epsilon/C

Assume |x+2|<1 so -1<x+2<1 = -3<x<-1 = -5<x-2<-3

After this step -5<x-2<-3, I'm inclined to set |x-2|<-3, but book states that |x-2|<5. I'm lost at this step, should it be|-5|? Anyways, any explanations would be great.

Thanks,
Chris

2. Oct 2, 2006

### Galileo

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?

3. Oct 2, 2006

### quasar987

Small detail that is important:

You only want to prove that "|f(x)-L|<epsilon whenever 0<|x-a|<delta".

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

4. Oct 2, 2006

### David_Vancouver

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)

5. Oct 2, 2006

### David_Vancouver

Sorry this part:
if 0 < |x + 2| < 1 then,
-3 < x < -1
|x - 2| < 5
is a "side part"

6. Oct 3, 2006

### hmm?

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.

Last edited: Oct 3, 2006