.99999 repeating = 1

x= .999 repeating
10x = 9.999 repeating
10x-x = 9.999 repeating - .999 repeating
9x= 9
x=1

'Undefined' and 'can be any number' are two different things. 0^0, for example, is undefined because it has two different values by definition (0^x = 0, x^0 = 1). Infinity isn't a number, so things like infinity/infinity just simply aren't defined, though there isn't much difficulty evaluating 1^infinity - = 1*1*1*1*... = 1.

Let's just agree that it sucks.

1^infinity can = e

Proof:
some number "S" = lim(x->infinity) of (1+1/x)^x
(natural log) lnS = lim(x->infinity) ln(1+1/x)^x
(property of logs) lnS = lim(x->infinity) x*ln(1+1/x)
(x = 1/(1/x))lnS = lim(x->infinity) (ln(1+1/x))/(1/x)
(L'Hopital's rule)lnS = lim(x->infinity)((1/(1+1/x))*(-x^-2))/(-x^-2)
(reduce)lnS = lim(x->infinity)(1/(1+1/x))
(take limit)lnS = 1/(1+1/infinity)= 1/(1+0)=1
(exponentiate both sides) e^lnS = e^1
S = e

Degenerate forms can be many things even if it doesn't appear so. (Note: this does not prove the existence or value of e, it proves that the lim(x->infinity) of (1+1/x)^x = e)

>>83
You cannot raise 1 to the power of infinity, because infinity is not a number. Instead, you would ask what is the limit as n->infinity of 1^n. Clearly, 1^n = 1 for any integer n, so
lim(1^n,n->infinity)=lim(1,n->infinity)=1

FOr your response, 83, you never once demonstrated that 1^infinity = e. You seem to be assuming that you can freely interchange (1^infinity) and (1+1/infinity)^infinity, which you clearly cannot. As I discussed above, 1^infinity is simply impossible to calculate because infinity is not a number. Instead, you have to take the limit as x->infinity of 1^x, and I proved at the beginning that lim(1^x,x->infinity)=1.

It's not even a question of "can you do certain steps to get around the math." It's simply that you were wrong in your statement that "1^infinity can = e." It cannot. Ever. Ever. Ever.

Take some real analysis before saying any more. And if you have taken real analysis, either you did not pay attention or your professor sucked.

>>83
You idiot, "can equal" doesn't make any fucking sense. Equality is unique because it's transitive. It either IS EQUAL or IS NOT EQUAL. Take your "can equal" and shove it up your ass.

>>85
Oh what? I only just removed it from that very same place!

(direct substitution)lim(x->infinity)(1+1/x)^x = (1+1/infinity)^infinity
=(1+0)^infinity = 1^infinity
Because one (or for that matter any positive constant) over the "biggest number possible" (infinity) yields an infinitesimally small number that at the "point" (or limit) at infinity becomes zero for lim(n->infinity)(1/n). If you don't believe me, try to plug in really large number for the function 1/x. You will not achieve 0 until the "point" (or limit) at infinity. Plug in all the positive numbers you want, it will ALWAYS be a little more than 0 until the "point" (or limit) at infinity. Granted, you can not find this "point" on the real number line but the 1/x function has a horizontal asymptote that is only touched when you "reach" "infinity" (more commonly known as taking the limit as x approaches infinity) (lim(x->infinity)(1/x)) = 0 not something close to 0 but instead equals 0).

And yes it is "can equal" because 1^infinity is not a specific number just in the same manner that the variable "x" "can equal" (have a value equivalent to) 1, 2 or any other real number.

For instance 1^infinity = 169
some number "S" = lim(x->infinity) of (1+ln(169)/x)^x
(natural log) lnS = lim(x->infinity) ln(1+ln(169)/x)^x
(property of logs) lnS = lim(x->infinity) x*ln(1+ln(169)/x)
(x = 1/(1/x))lnS = lim(x->infinity) (ln(1+ln(169)/x))/(1/x)
(L'Hopital's rule)lnS = lim(x->infinity)(-ln(169)/(x^2+x*ln(169)))/(-1/x^2)
(reduce, 1st step)lnS = lim(x->infinity)(x^2*ln(169))/(x^2+x*ln(169))
(reduce, 2nd step)lnS = lim(x->infinity)(x*ln(169)/(x+ln(169))
(add and subtract the same quantity, 3rd step)lnS = lim(x->infinity)(x*2*ln(13)+4*(ln(13))^2-4*(ln(13))^2)/(x+2*ln(13))
(seperate)lnS = lim(x->infinity)((x*2*ln(13)+4*(ln(13))^2)/(x+2*ln(13))-(4*(ln(13)^2)/(x+2*ln(13))
(factor)lnS = lim(x->infinity)(2*ln(13)*(x+2*ln(13))/(x+2*ln(13))-(4*(ln(13)^2)/(x+2*ln(13)))
(reduce, 3rd step)lnS = lim(x->infinity)(2*ln(13)-(4*(ln(13)^2))/(x+2*ln(13)))
(take limit)lnS = 2*ln(13)-4*(ln(13))^2/(infinity+2*ln(13))=2*ln(13)-0=ln(169)
(exponentiate both sides) e^lnS = e^ln(169)
S = 169

As you can see, the degenerate form 1^infinity can be ANYTHING. It's just a little bit harder to do those that aren't common exponents of e.

some number "S" = lim(x->infinity) of (1+ln(169)/x)^x
(natural log) lnS = lim(x->infinity) ln(1+ln(169)/x)^x

fail

\ln S = \ln \lim_{x \tendsto \inf} (1+\frac{\ln 169}{x})^x

actually you CAN use infinitesimals in math, but there are none in the real number set.

http://en.wikipedia.org/wiki/Infinitesimal

"0.000...1" is just a nonsense number, you can't have anything after an infinite number of decimal places by definition.

>>89
Someone needs to read http://en.wikipedia.org/wiki/Archimedean_property

>>90
Yes, you.

>>91
That be useless since i did read it already.I wouldn't point out articles that i didn't read myself.
The article describes how it defines real number by excluding infinitesimals
from the set.Archimedes himself used them http://en.wikipedia.org/wiki/Archimedes%27s_use_of_infinitesimals
and his disbelief in them helped shape math theory.
http://en.wikipedia.org/wiki/Infinitesimal s  are part of http://en.wikipedia.org/wiki/Hyperreal_number s which by no means
"non-existant" or absurd.

proof:

[9*(sigma from 1 to inf) 10^-x converges to 1].  This may seem like an approx, but it is not.  When an infinite series converges to a number, it IS that exact number.  This concept is used in calculus all the time for finding area under curves.  By saying that 0.999~ is not = 1, you are basically saying that integral calculus is useless (because integration is based on summing up an inf number of rectangles, kind of like the inf series shown above).

>>93
no yoor not

In other words >>93 tried to say:
proof that .999...=1:

.9999=1 QED!!111

>>95
Stop strawmanning or GTFO

>>96
TITS

Put .999... on a chart and youll never get to point 1. End.
Algebra II teaches this simple concept.

>>98
Not a sequence. Go away.

100 GET!

>>98
The limit of the sum that produces the decimal expansion that is .999... is 1, and therefore they are one and the same. High school calculus teaches this simple concept, and calc trumps Alg II any day.

Duh, the limit is 1

wow my first 100 topic

I've ejaculated over this thread four times already. Keep it coming!

dude... seriously who gives a shit

>>105
legions of high school kids who are smart enough to use intuition but too stupid to understand advanced math.

loooool

bump for justice

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