Precise definition of a limit


by hmm?
Tags: definition, limit, precise
hmm?
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#1
Oct1-06, 09:10 PM
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|<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
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Galileo
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#2
Oct2-06, 04:07 AM
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Quote Quote by hmm?
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|?
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?
quasar987
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#3
Oct2-06, 09:51 AM
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Small detail that is important:

Prove that lim x->-2 (x^2-1)=3

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

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

David_Vancouver
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#4
Oct2-06, 09:50 PM
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)
David_Vancouver
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#5
Oct2-06, 09:55 PM
P: 4
Sorry this part:
if 0 < |x + 2| < 1 then,
-3 < x < -1
|x - 2| < 5
is a "side part"
hmm?
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#6
Oct3-06, 12:10 AM
P: 19
Quote Quote by 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.
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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