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Quote from TheFieldZy »Nobody's perfect, so neither is Hannah Montana Linux, but it's pretty great.

Quote from BC_Programming on Operating Systems »They all suck. They just suck differently. Sort of like prostitutes.

something when that something has no actual value

since it carries on infinitely?

multiply both sides by 9...

9/9=.9 repeating

9/9 is also obviously equal to 1

so .9 repeating equals 1

It does have an actual value. That actual value is 1.

Infinity is no barrier to something having a well-defined value.

Now, can we not have this same argument again in a topic that's nearly a year old?

Already, this can only work if xdifferent from 0 (can't divide by 0)

what is x^x?

x^x=exp(x*ln(x)) =B

1/x=exp(-ln(x)) = A

A*B= (x^x)/x= exp(x*ln(x)-ln(x)) = exp(ln(x) (x-1)) ( factorization)

A*B=(exp(ln(x))^(x-1) (that is because exp(x)=e^x, so exp (nx)= e^(nx)=((e^x)^n)

And A*B = exp(ln(x)) = x

So A*B= x^(x-1)

Q.E.D (quod erat demonstrandum)

Now, for the limit in 0:

lim(x*ln(x)) in O = O

so lim(exp(x*ln(x))) in 0=1

so so exp x*ln(x) equivalent to 1 in 0

In 0 A*B equivalent to 1/x, and that limits towards infinity.