Addendum

e^x → 1 + x (as x → 0).


Because log e = 1, this could be written as.

e^x → 1 + x(log e) as x → 0.


Interestingly if we now generalise for n^x,

n^x → 1 + x(log n) as x → 0.


This therefore provides a ready means of approximating the value of log n (for any value of n).

So log n → (n^x - 1)/x as x → 0.


So for example if we set x = .000001,

log 2 → (2^.000001 - 1)/.000001 = (1.000000693147... - 1)/.000001 = .693147...


So this approximation is already correct to 6 decimal places with respect to its true value and the relative accuracy of the approximation can be continually increased through taking a smaller value of x.