Date: Sep 25, 2013 2:08 PM
Author: jeremy
Subject: the lim of h->0 { (a^h - 1) / h }  ==  ln a ?

I found this while trying to find the dy/dx(a^x)

so dy/dx (a^x)

= lim h->0 { [ a^(x+h) - a^x ] / h }

= lim h->0 { [ a^x * a^h - a^x ] / h }

= lim h->0 { [ a^x * ( a^h - 1 ) / h }

= a^x * lim h->0 { (a^h - 1) / h }

In fact, I know from a different way of proofing, dy/dx ( a^x ) = a^x * ln a,

It seems "ln a" is equivalent to "lim h-> 0 { (a^h - 1) / h}", but if I didn't know the latter proof, how can I solve the limit?