I started it because of that:
Thus, the unsigned infinity is the reciprocal of zero.
I just wanted to say, that the reciprocal of infinity isnt "0" unless you use limits or rather you just can express it with limits in a other way than 1/∞.
To use and undefinable value and then apply Limit to it is pointless.
So thats not the case....
Regardless of whether it makes sense to use it here or not my statement is correct.
lim 1/n = 0
n->∞
If you dont trust me just look wiki; near infinities; first case:
http://en.wikipedia.org/wiki/List_of_limitsNo matter what you divide 1 by even to an infinite amount it aproaches 0 but never reaches it.
Yes and a limit is a function to express this approaches, so its "0".
Following your detail
1/n = 0
(1/n)*n = 0*n
n/n=0
1=0
Which is implausible
If you operate just with "1/n = 0" it will automatically become invalid by ignoring "lim" and "n->∞".
http://en.wikipedia.org/wiki/Limit_(mathematics)#Limit_of_a_function
Shows a good example of using limit to define a theoretical value. We know that f(1.0) is equidistant between f(x) equidistant from 1 but that point of evaluation is undefineable so they have used a Lim to return the value 2.
I already know this example... its (x+1) expanded with (x-1)/(x-1), thus the term will be undefined with "x=1". To get the limit there you replace the "1" (or rather x) with an expression that approaches "1". For example lim(1 + 1/n); n->∞.
((x+1)(x-1))/(x-1)
replace "x" with lim(1 + 1/n); n->∞.
lim(((1 + 1/n)+1)*((1 + 1/n)-1))/((1 + 1/n)-1)=2
n->∞
(1+1) * reduced to "1"=2