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How come 1^(infinity) isn't 1?

it comes out to 1 anyway, its just you need to do your limit a different way. usually you get a 1^inf when you have a limit as n->inf of (n / n+1)^n. As n gets big the 1 doesn't matter and you get 1^inf. The way you solve that problem is you say that (n/n+1)^n = e^ln(n/n+1)^n, then move the n exponent down (log properties), giving you e^n*ln(n/n+1), do the limit and get 0 for the ln function, so you have e^(inf * 0) or e^0, which is 1.
 
But the real question is. Why do you have nothing better in your life right now to talk about other then math on an online forum? And why do I have the time to mock you?
 
Originally posted by: TallBill
But the real question is. Why do you have nothing better in your life right now to talk about other then math on an online forum? And why do I have the time to mock you?

math is fun
 
Originally posted by: TallBill
But the real question is. Why do you have nothing better in your life right now to talk about other then math on an online forum? And why do I have the time to mock you?

kinda like posting on AT right?
 
Originally posted by: Soccerman06
Originally posted by: TallBill
But the real question is. Why do you have nothing better in your life right now to talk about other then math on an online forum? And why do I have the time to mock you?

kinda like posting on AT right?

My point exactly.
 
Originally posted by: mercanucaribe
I don't get it. No matter how many times you multiply 1 by 1, it won't be anything but 1.

Most theorems/proofs have the domain of all real numbers. But infinity isn't a real number. It's more of a concept that we'll never understand.
 
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