So you’re saying that 0.99999… = 1?
I think this assertion calls for further discussion 
Only for small values of 1.
I could swear I had already made these corrections, but at any rate, here they are now, in bold.
Shit, another typo:
When is e day?
Feb. 72nd
a day each for MDY and DMY types.
Here’s a presentationI did that shows how Pi could be calculated using polygons.
You could therefore say the area of a circle is the sum of the areas of an infinite amount of polygons. Each polygon would have a rational number for its area, but since there’s an infinite amount of them, all those rational numbers add up to an irrational number.
Nice. I like your “Near Infinity” - It reminds me of the Classic Trek (“Wolf in the Fold”) where Spock, in order to keep the computer busy, orders it to calculate π to the last digit.
OMG. 
You owe me a new keyboard…
Another way to visualize pi that doesn’t rely on circles or polygons: pi is 4 times the area under the curve y = 1 / (1 + x[sup]2[/sup]) between x = 0 and x = 1.
The hidden connexion to circles here is that the integral of that curve is the arctangent. You can use that integral to calculate pi, although it converges very slowly so no one ever does.
Au contraire. One of the fastest way people compute π is by integrating 1/(1 + x[sup]2[/sup]). They just do it over different ranges and combine the result.
In more detail: 1/(1 + x[sup]2[/sup]) = 1 - x[sup]2[/sup] + x[sup]4[/sup] - x[sup]6[/sup] + … for |x| < 1, by the standard geometric series expansion, and thus its integral from 0 to x [arctan(x)] in that range is x - x[sup]3[/sup]/3 + x[sup]5[/sup]/5 - x[sup]7[/sup]/7 + … Put another way, its integral from 0 to 1/x [arccot(x)] is 1/x - 1/(3x[sup]3[/sup]) + 1/(5x[sup]5[/sup]) - 1/(7x[sup]7[/sup]) + …
This gives a very efficient means of computing arccot(x) for x > 1 [note that the terms in the series alternate between positive and negative but get smaller in magnitude by a factor of at least x[sup]2[/sup] each time; thus, the partial sums alternate between a sequence of lower upper bounds and a sequence of increasing lower bounds, with the distance between these shrinking in accordance with the size of the terms].
[We can also compute arccot(1) directly with this series (which, for x = 1, would be 1 - 1/3 + 1/5 - 1/7 + …), but very slowly (now the shrinking by at least a factor of x[sup]2[/sup] is of no use; here, instead, convergence comes from the fact that 1/n goes to 0, but much more slowly).]
So instead of computing π/4 as arccot(1), we can use representations of arccot(1) as combinations of arccot(n) for larger n [“Machin-like formulae”]. The simplest of these is arccot(1) = arccot(2) + arccot(3) [evidenced by the observation that (2 + i) * (3 + i) = 5 * (1 + i)], and this gives a method of computing π which is already very efficient. World record computations are often done by the same technique, just using slightly more honed decompositions [e.g., arccot(1) = 12 arccot(49) + 32 arccot(57) - 5 arccot(239) + arccot(110443), again verifiable by simple complex number calculation as before].