1. ## Solution: epsilon proof: lim 0.9999...=1

I see more in your proof that is wrong then is right. For example equation (1) is actually equal to 2. Equation (2) you are claiming that 2 < 1.something, so is also wrong. In equation 3 you lost a 9 somewhere.

Your proof is so riddled with notation and math errors that it is pretty much meaningless.
Posted Jul25-12 at 01:36 PM by Integral
2. ## Laplace Transform Via Limits

There's no need for voodoo. We just take the n limit first to get a function of m and then take the m limit.
$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m/n) \right)^n}{1-exp(-(s+ \alpha )m/n)} \frac{m}{n} = \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{1-1+(s+ \alpha )m/n+O(n^{-2})} \frac{m}{n}$

$= \lim_{m->oo} \lim_{n->oo}m \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )m+O(n^{-1})} = \lim_{m->oo} \lim_{n->oo}\frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )+O(n^{-1})} = \lim_{m->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )} = \frac{1}{(s+ \alpha )}$
Provided that $(s+ \alpha )>$, which is exactly the same condition as is needed for the proper integral to have a finite value.
Posted Jul18-12 at 11:53 PM by andrewkirk