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**nbalakerskb****Member**- Registered: 2006-09-10
- Posts: 4

prove limx->2(3x-2)=4

that means i have to prove: ∣(3x-2)-4∣<ε whenever∣x-2∣<δ

next step: ∣(3x-6)∣<ε whenever∣x-2∣<δ

next step: ∣x-2∣<ε/3whenever∣x-2∣<δ

for the next step , we let

ε/3=δ

∣(3x-2)-4∣=∣(3x-6)∣=3∣x-2∣<3δ=3(ε/3)=ε

thus ∣(3x-2)-4∣<ε whenever∣x-2∣<δ

is that correct? um... coz i 'm not really understand why we need to let ε/3=δ

thx for help

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

ehm, lim of x to what? to infintity? to 0, to -infinty? what?

or do you mean

in which case, wouldnt this be satisfactory

lim(x->2) 3x = 3*2 = 6, -2 = 4 ?

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**nbalakerskb****Member**- Registered: 2006-09-10
- Posts: 4

luca-deltodesco wrote:

ehm, lim of x to what? to infintity? to 0, to -infinty? what?

or do you mean

in which case, wouldnt this be satisfactory

lim(x->2) 3x = 3*2 = 6, -2 = 4 ?

i mean lim of x to 2 and the function is (3x-2)

thx

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

when you have a constant like that, you can basicly just ignore it, work out the limit for anything involving the variable, and then do what needs done with the constant. i.e.

etc etc.

so in this case, obviously, as x approaches 2, 3x approaches 6, take away 2, and you get 4.

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):

Show that x² --> 4 when x --> 2.

We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!

|x²-4| = |x+2|*|x-2|

Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5. If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:

|x²-4| = |x+2|*|x-2|<5*ε/5=ε

Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...

That's all there is, I think. I don't see where you got that ε/3=δ from though?

*Last edited by numen (2006-09-10 10:09:11)*

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**nbalakerskb****Member**- Registered: 2006-09-10
- Posts: 4

numen wrote:

I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):

Show that x² --> 4 when x --> 2.

We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!|x²-4| = |x+2|*|x-2|

Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5.If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:|x²-4| = |x+2|*|x-2|<5*ε/5=ε

Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...

That's all there is, I think. I don't see where you got that ε/3=δ from though?

i don't really understand this part... why we have to demand |x-2|<1...

and why the |x+2| <5 can lead the next step:|x-2|<ε/5 ? is it some kind of basic inequality properties?

i am confused... thx for help anyway

*Last edited by nbalakerskb (2006-09-10 14:00:15)*

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