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Topic: the lim of h->0 { (a^h - 1) / h } == ln a ?
Replies: 1   Last Post: Oct 6, 2013 3:14 PM

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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
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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?



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