Epsilon Delta Limit.

ƒ(x)

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

tiny-tim

hi ƒ(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 )

ƒ(x)

Can you explain yourself a little further? I tend to be rather slow about this kind of thing.

tiny-tim

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

ƒ(x)

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

hi ƒ(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?

you just have to guess …

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