# Prove by definition a limit

[PLAIN]http://img408.imageshack.us/img408/503/limesn.jpg [Broken]

I'll be most thankful for any help :)

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Dick
Homework Helper
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.

Can someone perhaps show me the deltas that apply? I'm not sure i'm doing it absolutely right.

HallsofIvy
Homework Helper
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.

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?

Dick
Homework Helper
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.

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)?

Dick