Prove by definition a limit

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  • #1
gipc
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Can someone please help me prove the following limit using epsilons and deltas?
[PLAIN]http://img408.imageshack.us/img408/503/limesn.jpg [Broken]

I'll be most thankful for any help :)
 
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  • #2
Dick
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Split the problem up. Find a delta for the rationals, and then find a delta for the irrationals. Then just take the minimum of the two deltas. It's just a combination of two problems you probably already know how to do.
 
  • #3
gipc
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Can someone perhaps show me the deltas that apply? I'm not sure i'm doing it absolutely right.
 
  • #4
HallsofIvy
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If you are not sure you are doing it absolutely right, then, hopefully, you are doing something! Show us what you are doing and we will try to help.
 
  • #5
gipc
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I'm unsure how to handle the fact the x belongs to Q or doesn't belong to Q and how it affects the solution. The problem is that I have to give a solid proof using the definition and I don't know how it changes things.

Is this correct?
lim x+2=1 as x->(-1)

I need to show that 0<|x+1|<delta => |f(x)-L|=|x+2-1|=|x+1|<?epsilon
so if I choose delta=epsilon the condition will stand.
is this okay?

and what about lim x^2 as x->(-1)
if 0<|x-1|<delta then i need to show that for every epsilon,
|x^2 -1|<epsilon
but |x^2-1|=|x-1||x+1|<|x+1|
so if I choose delta=epsilon+1 this will work.

Did i choose the correct deltas?
 
  • #6
Dick
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The first one is ok. For the second one you want |x^2-1|=|x-1|*|x+1|<epsilon. But |x-1|*|x+1| isn't less than |x+1|. Make sure delta is always less than 1, eventually say delta=min(1,...). Then since |x+1|<1, -2<x<0. Then |x-1| is between 1 and 3.
 
  • #7
gipc
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Let me see if I understand.
since |x+1|<1 then 1<|x-1|<3 and then we can see more clearly that |x^2-1|=|x-1||x+1|<|x+1| (because |x-1| is bigger than |x+1|).

so we get |x+1|<?epsilon
and thus delta=epsilon-1

so in the end we'll take delta=min(1, epsilon-1)?
 
  • #8
Dick
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|x-1||x+1| isn't less than |x+1|. I already told you this. If x=(-2), |x-1|=3. 3*|x+1| isn't less than |x+1|.
 

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