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Posted: 10/13/2004 7:49:38 PM EST
[Last Edit: 10/13/2004 7:49:56 PM EST by ColonelKlink]
Link Posted: 10/13/2004 7:53:50 PM EST
yes multiply by 4
Link Posted: 10/13/2004 7:55:12 PM EST

Originally Posted By ColonelKlink:
Im working with alternating infinite series shit for calculus three and I cant remember from calculus one why the limit of a negative integer with and exponent x that approaches infinity is undefined. I know it oscillates back and fourth so the limit doesnt exist, but does anyone know how to break it down mathematically to prove it?.

Thanks,
Klink.



We speak English in the US son!

And that right there is why I hate adv math...
Link Posted: 10/13/2004 7:56:36 PM EST
[Last Edit: 10/13/2004 7:58:46 PM EST by olyarms]
Edite, lol mis read, I am a dumb ass for give me. lease forgive lol
?
Link Posted: 10/13/2004 7:57:52 PM EST
Yes, mathematically speaking.


if you draw a graph of the function, and it represents asymtotes, then mathematically you could show that as each number approches the limit, but never touches it because 1 turns into .5, then .25, then one eigth and so on.


it gets so close but never touches the limit because the fraction just gets cut in *half each time.

hope this helps.
Link Posted: 10/13/2004 7:57:56 PM EST
[Last Edit: 10/13/2004 8:09:29 PM EST by topgunpilot20]
doesn't seem like it can be broken down anymore. A^inf=inf (therefore, the lim as x approached inf = inf)
Link Posted: 10/13/2004 8:01:22 PM EST
I was taught that if X=^5 then infinity is undefined...only if it appraoching X.

Link Posted: 10/13/2004 8:28:27 PM EST
[Last Edit: 10/13/2004 8:29:13 PM EST by ColonelKlink]
Link Posted: 10/13/2004 8:29:42 PM EST

Originally Posted By ColonelKlink:

Originally Posted By topgunpilot20:
doesn't seem like it can be broken down anymore. A^inf=inf (therefore, the lim as x approached inf = inf)


no what i am saying is that this
home.comcast.net/~colonelklink/lim.jpg

is undefined.







nevermind then.


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