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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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Posts: 132
Registered: 11/27/12
Re: the lim of h->0 { (a^h - 1) / h } == ln a ?
Posted: Oct 6, 2013 3:14 PM
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Saying that "lim_{h->0}{(a^h- 1)/h}= ln(a)" is the same as saying that, for sufficiently small h, (a^h- 1)/h is approximately equal to ln(a). That is the same as saying that a^h- 1 is approximately equal to hln(a)= ln(a^h).

In many texts "e", the base of the natural logarithms, is DEFINED by "lim_{n-> infinity} (1+ 1/n)^n". Replacing "1/n" by h, n going to infinity is the same as h going to 0 so "lim_{h-> 0} (1+ h)^{1/h}= e. That is, for h sufficiently small, (1+ h)^{1/h} is approximately equal to e so that 1+ h is approximately equal to e^h and then that h is approximately equal to e^h- 1 and, finally, that (e^h- 1)/h is approximately equal to 1. That is "lim_{h-> 0} (e^h- 1)/h= 1".

Of course, we can write a= e^{ln(a)} and so a^h= (e^{ln(a)})^h. Then a^h- 1= (e^{ln(a)})^h- 1= e^{hln(a)}- 1. Now, as h goes to 0, h ln(a) goes to 0 as well so replacing the "h" above with hln(a), "lim_{h->0} (e^{hln(a)}- 1))/h ln(a)= 1". But, of course, h ln(a)= ln(a^h) so that is equivalent to "lim_{h->0} (e^{ln(a^h)}- 1}{h ln(a)}= 1"

Finally, e^{ln(a^h)}= a^h and "lim_{h->0} (a^h- 1)/h ln(a)= 1" and, since ln(a) is independent of h, "lim_{h->0} (a^h- 1)/h= ln(a)[/tex].

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