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jeremy
Posts:
6
From:
santa clara
Registered:
9/25/13


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


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?



