Proof:
some number "S" = lim(x->infinity) of (1+1/x)^x
(natural log) lnS = lim(x->infinity) ln(1+1/x)^x
(property of logs) lnS = lim(x->infinity) x*ln(1+1/x)
(x = 1/(1/x))lnS = lim(x->infinity) (ln(1+1/x))/(1/x)
(L'Hopital's rule)lnS = lim(x->infinity)((1/(1+1/x))*(-x^-2))/(-x^-2)
(reduce)lnS = lim(x->infinity)(1/(1+1/x))
(take limit)lnS = 1/(1+1/infinity)= 1/(1+0)=1
(exponentiate both sides) e^lnS = e^1
S = e
Degenerate forms can be many things even if it doesn't appear so. (Note: this does not prove the existence or value of e, it proves that the lim(x->infinity) of (1+1/x)^x = e)