Without using L'Hopital's rule, prove that $$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$
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1What is your definition of $\ln a$? – Mark Bennet Nov 23 '15 at 08:54
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@MarkBennet: Natural logarithm – Mohamed Mostafa Nov 23 '15 at 08:55
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yes, but what is natural logarithm, how is it defined? – AlvinL Nov 23 '15 at 08:58
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2The natural logarithm can be defined - and is defined in some approaches - by the limit you are computing. Sometimes it is defined as an integral. All the definitions prove to be equivalent, but the point is, which definition are you using, because that affects how you might go about proving the statement. – Mark Bennet Nov 23 '15 at 09:40
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Don't know if that is rigorous enough for you, but this is one way using $e^t = \sum_{k=0}^{\infty} \frac{t^k}{k!}$, $e^{\ln x} = x$ and some results about absolute convergent series:
\begin{align*} \frac{a^x - 1}{x} & = (e^{x \cdot \ln a} -1)/x = \left(\sum_{k=0}^{\infty} \frac{(x \cdot \ln a)^k}{k!} - 1 \right)/x\\ & = \left( x \cdot \ln a + \frac{(x \cdot \ln a)^2}{2} + \ldots \right)/x = \ln a + (\ln a)^2 \cdot \frac{x}{2} + (\ln a)^3 \cdot \frac{x^2}{6} +\ldots\\ \end{align*}
Hence, $\lim_{x \to 0} \frac{a^x -1}{x} = \ln a$.
j4GGy
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May I suggest you write $x\cdot\log a$ instead of $\log a\cdot x$ ? – Claude Leibovici Nov 23 '15 at 09:12
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Thanks for doing it ! The initial writing could have been slightly confusing. Cheers :-) – Claude Leibovici Nov 23 '15 at 09:26