Looking over notes from class today and wanted to know if there is any type of proof for the fact that $\ln(a) = \lim_{h\to0}(a^h-1)/h$, which is just $f '(0)$ for any function of the form $f(x) = a^x$. I see that it simply is the case but where's the "mathy" proof of it? Thanks.
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Was $e^x$ defined to be the exponential whose slope at zero is 1? – Paul Sundheim Sep 29 '14 at 22:09
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Yes but for any other value it holds so where's the proof of that? For instance, if we had f(x)=3^x, then the slope at 0 for 3^x would be ln3. – King Squirrel Sep 29 '14 at 22:11
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1How is $a^x$ defined? – Daniel Fischer Sep 29 '14 at 22:12
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1Related. – Hakim Sep 29 '14 at 22:17
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an answer – Semsem Sep 29 '14 at 22:24
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@KingSquirrel The answer below that starts "Look at it this way" is a good proof. It requires that you know that the derivative of $e^x$ is itself which begins with knowing that the derivative of $e^x$ at zero is one. – Paul Sundheim Sep 29 '14 at 22:28
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@PaulSundheim, how so? It assumes what the derivative is and then proceeds to evaluate $f'(0)$. You might as well right that $f'(x) = \log(a) \cdot a^x$ implies $f'(0) = \log(a) \cdot 1 = \log(a)$ directly. The proof is invalid. – Christopher K Sep 29 '14 at 22:43
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The latest proof by Chris K works well and uses the starting point I was asking about and uses it to conclude the result you were looking for without the need to find the full derivative of $a^x$. The method I was suggesting works too, but goes through more steps. – Paul Sundheim Sep 29 '14 at 23:34
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Hint:
Use the definition $$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \\=\lim_{h\to 0}\frac{a^h-1}{h}\\=\lim_{h\to 0}\frac{e^{\ln a^h}-1}{h}\\=\lim_{h\to 0}\frac{e^{h\ln a}-1}{h}\\=\lim_{h\to 0}\frac{1+h\ln a+\cdots-1}{h}=\ln a$$
Semsem
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1To apply L'Hospital rule you need to know the derivative, and that is posted question. – mfl Sep 29 '14 at 22:15
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First use the commonly known limit "$\lim \frac{e^h-1}{h}$", that is:
$$1 = \lim_{x \rightarrow 0} \frac{e^x - 1}{x} \\ = \lim_{\log(a)x \rightarrow 0} \frac{e^{\log(a)\cdot x} - 1}{\log(a)\cdot x} \\ = \lim_{x \rightarrow 0} 1/\log(a)\cdot \frac{a^x - 1}{x}$$
and so $f'(0) = \log(a)$ by the definition of the derivative.
Christopher K
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Look at it this way: $$f(x)=a^x \implies f'(x)=a^x \ln a$$ this follows from the fact that of $y=a^x$ then $\ln y =x \ln a$ and so $1/y \cdot y' = \ln a \implies y'=a^x \ln a$. So we proved the derivative in other words (using the definition of $e$ as the exponential function base which has a derivative equal to the function itself).
There is another way to go about finding this derivative, which is to use the definition of the derivative: $$f(x)=a^x \implies f'(x)= \lim_{h\rightarrow 0} \frac{a^{x+h}-a^x}{h}=\lim_{h \rightarrow 0} \frac{a^x(a^h-1)}{h}=a^x \cdot \lim_{h \rightarrow 0} \frac{a^h-1}{h}$$ Now set the two derivatives equal to one another: $$a^x \ln a = a^x \cdot \lim_{h \rightarrow 0} \frac{a^h-1}{h} \implies \ln a= \lim_{h \rightarrow 0} \frac{a^h-1}{h}$$
Sheheryar Zaidi
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