Epsilon-delta definition

[spongebob]

Dammit, I said it's really a white dry-erase board!! I just call it a chalk board 'cuz I'm sick of people asking what a dry-erase board is.




actually, I have two of them. one huge two-sided flip-board, and one wall unit.

well you could have been up front with me, i use dry erase boards at work.




__________________________________________________

cereal killer!

sponge is no longer associated with the associates association.

Titanium Member

 

[spongebob]

I saw this and was like, fuck it, I don't have time to do that shit, but then I was wondering what the solution was, so I just did it.

God, I'm pathetic.

curious sammy!




__________________________________________________

cereal killer!

sponge is no longer associated with the associates association.

Titanium Member

 

[Spartacus]

I was getting all this for my homework problems then the teacher gave us these problems and now I am lost... Grrr

 

[courtneybcca]

I saw this and was like, fuck it, I don't have time to do that shit, but then I was wondering what the solution was, so I just did it.

God, I'm pathetic.

goodness you always come to my rescue thank you so much
MUAH!!!!

 

[samoth]

goodness you always come to my resuce thank you so much
MUAH!!!!

I thought this was a failed attempt at pwning Samoth? I'm confused. Either way, no one ever asked for the solution, so I'm assuming it was a joke.

I hate you all.

 

[courtneybcca]

no it is not a joke... dont listen to everhung he has a immature crush on me and if he doesn't stop fucking with my posts I will get him kicked off

the exact question is to prove the limit using the epsilon delta defintion
I just used the graphical method it to see it

 

[Spartacus]

crush?
heh

 

[samoth]

Okay, first of all proof will require to show that your δ that you arrive at works when put back in the equation.

You found a number δ such that

|(-x³+1)+7|<ε whenever 0<|x-2|< δ

Since |(-x³+1)+7|=|(-x³+8)|= |(x-2)||(x²+2x+4)|,

You want |(x-2)||(x²+2x+4)|<ε when 0<|x-2|< δ

This gives you |x-2|< ε/|(x²+2x+4)| when 0<|x-2|< δ

So you propose that δ= ε/|(x²+2x+4)|

For the proof, show that this specific value of δ works.
Given ε>0, choose δ= ε/|(x²+2x+4)|

Then if 0<|x-2|< δ,

|(-x³+1)+7|=|(-x³+8)|= |(x-2)||(x²+2x+4)| < |(x²+2x+4)| δ

=(|(x²+2x+4)|) (ε/|(x²+2x+4)|) = ε


SO

|(-x³+1)+7|<ε whenever 0<|x-2|< δ

Thus by the definition of the limit,
Lim (-x³+1) = -7
x→2

Q.E.D.



Lemme know if I made a mistake or somethin'. I only used numerical methods and algebreic manipulations, I didn't graph anything. But I don't think answering it should be dependent on that, since the proof won't be.

 

[Spartacus]

I miss all the tits and ass in college

 

[samoth]

I miss all the tits and ass in college

Don't feel bad, so do I.