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Epsilon Delta Limit.

by (x)
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(x)
#1
Feb15-12, 05:21 PM
P: 323
For part A, (described here: http://www.cramster.com/solution/solution/1157440) I don't understand why they say |x-2| < 1 and why [itex]\delta[/itex] = min{1,ε/5}

In case you can't view the page:

lim x2+2x-5 = 3, x [itex]\rightarrow[/itex] 2
Let ε > 0 and L = 3.

|x2 + 2x -5 -3| < ε
|x2 + 2x - 8| < ε
|x+4||x-2| < ε

If |x-2| < 1, x [itex]\rightarrow[/itex] (1,3) [itex]\rightarrow[/itex] x+2 [itex]\in[/itex] (3,5)
|x-2| < 1 [itex]\rightarrow[/itex] |x+2| < 5
|x-2| < 1 [itex]\rightarrow[/itex] |x2 +2x - 8| = |x-4||x+2| < 5|x-4|

For ε > 0, 5|x-4| < ε [itex]\rightarrow[/itex] |x-4| < [itex]\frac{ε}{5}[/itex]

If [itex]\delta[/itex] = min{1,[itex]\frac{ε}{5}[/itex]}
|x-2| < [itex]\delta[/itex] [itex]\rightarrow[/itex] |x2 + 2x - 8| < 5|x-4| < ε
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tiny-tim
#2
Feb15-12, 05:38 PM
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hi (x)!
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I don't understand why they say |x-2| < 1 and why [itex]\delta[/itex] = min{1,ε/5}

|x+4||x-2| < ε
because if δ < 1, then |x+4| < 5

and if δ < ε/5, then |x-2| < ε/5 (obviously!)
so |x+4||x-2| < 5*ε/5 = ε
(though i don't see why they say |x-2| < 1 either )
(x)
#3
Feb15-12, 05:40 PM
P: 323
Quote Quote by tiny-tim View Post
hi (x)!


because if δ < 1, then |x+4| < 5

and if δ < ε/5, then |x-2| < ε/5 (obviously!)
so |x+4||x-2| < 5*ε/5 = ε
(though i don't see why they say |x-2| < 1 either )
Can you explain yourself a little further? I tend to be rather slow about this kind of thing.

tiny-tim
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Feb15-12, 05:47 PM
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Epsilon Delta Limit.

Quote Quote by (x) View Post
Can you explain yourself a little further? I tend to be rather slow about this kind of thing.
you want the product to be < ε

one easy way to do this (there are others) is to make one of them < 1 and the other < ε

but that won't work in this case, since |4+x| isn't going to be < 1
however, it will be < 5,

so we make one of them < 5 and the other < ε/5
(x)
#5
Feb15-12, 06:03 PM
P: 323
Quote Quote by tiny-tim View Post
you want the product to be < ε

one easy way to do this (there are others) is to make one of them < 1 and the other < ε

but that won't work in this case, since |4+x| isn't going to be < 1
however, it will be < 5,

so we make one of them < 5 and the other < ε/5
So, for a different problem, part (c) on the link, I have:

lim x^3 + 2x + 1 = 4, x --> 1

Let ε > and L = 4

|x-1| < δ, |x^3 +2x - 3| < ε

But this one doesn't factor...[EDIT] wait, yes it does (I peeked at the solution, how do I factor this by myself?).

|x-1||x^2 + x + 3| < ε
tiny-tim
#6
Feb16-12, 04:05 AM
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hi (x)!

(just got up )

ok, so |x-1| < δ,

and you can obviously choose x near 4 so that |x2 + x + 3| < 33
so how would you finish the proof?
Quote Quote by (x) View Post
But this one doesn't factor...[EDIT] wait, yes it does (I peeked at the solution, how do I factor this by myself?).
you just have to guess

in an exam, they won't give you anything difficult, so start with 1, and work your way upwards!


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