Epsilon-delta definition

[courtneybcca]

Okay I need some help I think I got the answer but not sure

prove the lim as x approach 2 (-x^3 +1)=-7 using the epsilon-delta definition of the limit

so for every epsilon greater than zero there exists a delta greater than zero

0<lx-2l<delta then, l(-x^3 +1) +7l<epsilon

which works out to l-x^3 +8l<epsilon

when you do the graph I am getting delta: 1 as 2- cubed root of (8-epsilon) and delta: 2 as cube root of (8+epsilon) -2

Not sure how this relates
I was getting all this for my homework problems then the teacher gave us these problems and now I am lost... Grrr

 

[slickdadd]

I might be able to help if you explained the epsilon delta method in a bit more detail. Not sure how that works.

 

[spongebob]

you need to post this in the thread titled 'crisis the sixth'.

sammy the math wizard is a post whore in there, i dont even think he has viewed any other threads lately. he has his own blackboard in his dorm.




__________________________________________________

cereal killer!

sponge is no longer associated with the associates association.

Titanium Member

 

[courtneybcca]

I might be able to help if you explained the epsilon delta method in a bit more detail. Not sure how that works.

this is the basics of calculus if you dont already know this there is no way you can help me

 

[slickdadd]

this is the basics of calculus if you dont already know this there is no way you can help me

I haven't had basic calculus in a few years therefore I don't remember the theorems off the top of my head. But, suit your self....

 

[courtneybcca]

I totally appreciate you trying to help but I dont know it well enough to explain...

 

[samoth]

See VI.

 

[samoth]

sammy the math wizard is a post whore in there, i dont even think he has viewed any other threads lately. he has his own blackboard in his dorm.

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.

 

[spongebob]

see courtney, post this shit in the other thread. sammy will knock it out in no time flat.




__________________________________________________

cereal killer!

sponge is no longer associated with the associates association.

Titanium Member

 

[samoth]

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.

 

[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.

 

[Spartacus]

Don't feel bad, so do I.

now that was good

 

[samoth]

now that was good

 

[Yarg!]

hey samoth or courtney can u guys solve this for me please??
two ships leave port at the same time. one sails south at 15mi/h and the other sails east at 20m/h. find a function that models the distance D between ships in terms of the time t (in hrs) elasped since their departure. thanks.

 

[courtneybcca]

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.

Okay so wow I totally understand the way you did that... But that is way more advanced then we have done in are class yet... Why would are teacher give us something like this that we haven't done yet.. That pisses me off any way though thank you so much!!! Do you happen to have msn or yahoo messenger... If so could you please pm me with the address....
Thank you so much

 

[Spartacus]

are should be our
go fuck yourself

 

[samoth]

Okay so wow I totally understand the way you did that... But that is way more advanced then we have done in are class yet... Why would are teacher give us something like this that we haven't done yet.. That pisses me off any way though thank you so much!!! Do you happen to have msn or yahoo messenger... If so could you please pm me with the address....
Thank you so much

It happens quite often in more areas than just calc. Sometimes things are "beyone the scope of the class", which I've been hearin' for the past 4 years, lol, and getting too deeply into something just confuses *most* people or raises too many questions. Or the prof/TA feels that a certain way of progression through the material would be better.

Also, some profs and some texts just plain suck. That's why I never go by just one source of course material anymore.

For the proof I wrote, I used the exact equation given in Stewart's Calculus Early Transcnedentals 4E (section 2.4 pg.115). It was only a matter of plug and chug. You might want to check out some supplementary materials for the course. Or you guys may just do things differently, and your prof/TA doesn't think some things are relevant or would just bog most people down.

I've never used MSN or Yahoo Messenger in my life, but I can be reached at Purdue#NoSpam.Hushmail.com if you would like. I'm sure you can figure the addy out. If not, PM me.

 

[Yarg!]

guys plz help me out anytime youre ready is fine. i fucking hate these word problems. im glad im not at ur guys lvl though.

 

[samoth]

guys plz help me out anytime youre ready is fine. i fucking hate these word problems. im glad im not at ur guys lvl though.

You have speed and direction: you're dealing with vectors. Draw the problem out. Draw the metric between the heads of the two vectors -- this is your distance. It varies as a function of time with respect to the two ships.

Key: Think Pathagorem's triangle. D(t)=sqrt(v1(t)^2 + v2(t)^2). Remember to label your axes correctly, as you're working in the cartesian coordiate system.

 

[Spartacus]

cut it out samoth
get an erection
rub it out
that's my formula