1^infinity=Q

lim(x->infinity)(1+Q/x)^x, Q is some positive real number
direct Substitution yields: (1+Q/infinity)^infinity=(1+0)^infinity=1^infinity
because lim(n->infinity)(1/n)=0
then let some number "S"=lim(x->infinity)(1+Q/x)^x
natural log: lnS=lim(x->infinity)ln(1+Q/x)^x
because compound function limit properties: g(lim(x->infinity)f(x))=lim(x->infinity)g(f(x))
property of natural logs:lnS=lim(x->infinity)x*ln(1+Q/x)
x=1/(1/x):lnS=lim(x->infinity)ln(1+Q/x)/(1/x)
L'Hopital's rule:lnS=lim(x->infinity)(-Q/(x^2+Q*x))/(-1/x^2)
reduce:lnS=lim(x->infinity)(Q*x)/(Q+x)
(brief digression)some expression "A" fulfills the following equation: (Q(Q+x))/(Q+x)+A/(Q+X)=(Q*x)/(Q+x)
multiply by (Q+x) on both sides: Q^2+Q*x+A=Q*x
subtract (Q^2+Q*x) from sides:A=Q*x-Q*x-Q^2
reduce:A=-Q^2
substitution:lnS=lim(x->infinity)(Q(Q+x))/(Q+x)-Q^2/(Q+x)
reduce:lnS=lim(x->infinity)(Q-Q^2/(Q+x))
take limit:lnS=Q-Q^2/(Q+infinity)
reduce:lnS=Q-0=Q
raise to e on both sides: S = Q

I'll explain any part that sounds confusing. (though I'll refer back to previous posts if people ask the same question)

It would except for the fact that infinity/infinity has more than one answer and most of those answers are distinct (one that isn't is lim(x->infinity)ln(x)/x Vs. lim(x->infinity)x/x^2 (both are 0)). Again, infinity does not play by the normal rules. Infinity does not have to equal any ONE thing. Infinity is not part of the standard real numberline. Also, because Infinity equals different things under different circumstances, all reals stay their respective values since no function can have two values as you approach infinity for that function.