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Epsilon Delta Limit. 
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#1
Feb1512, 05:21 PM

P: 323

For part A, (described here: http://www.cramster.com/solution/solution/1157440) I don't understand why they say x2 < 1 and why [itex]\delta[/itex] = min{1,ε/5}
In case you can't view the page: lim x^{2}+2x5 = 3, x [itex]\rightarrow[/itex] 2 Let ε > 0 and L = 3. x^{2} + 2x 5 3 < ε x^{2} + 2x  8 < ε x+4x2 < ε If x2 < 1, x [itex]\rightarrow[/itex] (1,3) [itex]\rightarrow[/itex] x+2 [itex]\in[/itex] (3,5) x2 < 1 [itex]\rightarrow[/itex] x+2 < 5 x2 < 1 [itex]\rightarrow[/itex] x^{2} +2x  8 = x4x+2 < 5x4 For ε > 0, 5x4 < ε [itex]\rightarrow[/itex] x4 < [itex]\frac{ε}{5}[/itex] If [itex]\delta[/itex] = min{1,[itex]\frac{ε}{5}[/itex]} x2 < [itex]\delta[/itex] [itex]\rightarrow[/itex] x^{2} + 2x  8 < 5x4 < ε 


#2
Feb1512, 05:38 PM

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hi ƒ(x)!
and if δ < ε/5, then x2 < ε/5 (obviously!) so x+4x2 < 5*ε/5 = ε(though i don't see why they say x2 < 1 either ) 


#3
Feb1512, 05:40 PM

P: 323




#4
Feb1512, 05:47 PM

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Epsilon Delta Limit.
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, 


#5
Feb1512, 06:03 PM

P: 323

lim x^3 + 2x + 1 = 4, x > 1 Let ε > and L = 4 x1 < 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?). x1x^2 + x + 3 < ε 


#6
Feb1612, 04:05 AM

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hi ƒ(x)!
(just got up …) ok, so x1 < δ, and you can obviously choose x near 4 so that x^{2} + x + 3 < 33 … so how would you finish the proof? in an exam, they won't give you anything difficult, so start with ±1, and work your way upwards! 


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