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I was going through and old book and I came across with a discussion which culminated with something like

\begin{align*} \ln(x)=\lim_{n\to \infty} 2^n\cdot (x^{1/2^n}-1) \end{align*} for every $x>0$. Is it really true? If that is the case, how to prove it?

I tried to test some values using using Geogebra and Google Sheets and both expressions produced really close values for small $n$ but it became impractible to analyse the behaviour of $2^n\cdot (x^{1/2^n}-1)$ as $n$ goes to infity for $2^n$ grows too fast.

Thanks.

PtF
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    You can identify $2^{-n}$ as $h$ and then arrive at something that is probably more familiar to you. – Ian Sep 08 '22 at 22:10
  • Yep, that's right! Now I get the point. The author wants to define $ln(x)$ using the above expression, that is not the case that he tries to show it. He then proceeds to show all the usual properties of $\ln$. He then uses this to define $e$, $e^x$ and $a^x$. That is an interesting approach. – PtF Sep 08 '22 at 22:12
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    This has been asked a few times already, e.g. at Logarithm approximation. – Qiaochu Yuan Sep 08 '22 at 23:14

2 Answers2

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We can let $t=1/2^n$, and our limit becomes $$\lim_{n\to\infty}2^n\left(x^{1/2^n}-1\right)=\lim_{t\to0}\frac1t\left(x^t-1\right).$$ Then, we can rewrite this as $$\lim_{t\to0}\frac{x^t-1}t=\lim_{t\to0}\frac{x^t-x^0}{t-0}=\left.\left.\frac{d}{dt}x^t\right|_{t=0}=\log x\cdot x^t\right|_{t=0}=\log x$$ as desired.

Andrew Chin
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  • ...unless for some reason the derivative of the exponential function has not yet been learned. – Andrew Chin Sep 08 '22 at 22:27
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    It should be $t \to 0^+$ instead of $t \to 0$. – jjagmath Sep 08 '22 at 22:54
  • @jjagmath Does this then invalidate the whole thing as I've only found a one-sided limit? I've found the one-sided derivative? – Andrew Chin Sep 08 '22 at 23:04
  • No. You actually found the (two-sided) limit $\lim_{t\to0}\frac{x^t-1}t = \log x$ which implies the one-sided limit $\lim_{t\to0^+}\frac{x^t-1}t $ is also equal to $\log x$. Just need to be careful because that won't always be the case. – jjagmath Sep 08 '22 at 23:08
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Yes, $$\lim_{n \rightarrow \infty} 2^n(x^{2^{-n}}-1) = \ln x.$$

HINT: Replace $2^n=Y$. Then $$\lim_{n \rightarrow \infty} 2^n(x^{2^{-n}}-1) \ = \ \lim_{Y \rightarrow \infty} Y(x^{\frac{1}{Y}}-1).$$

Then $$\lim_{Y \rightarrow \infty} Y(x^{\frac{1}{Y}}-1) = \lim_{Y \rightarrow \infty} Y \times (e^{\frac{\ln x}{Y}}-1)$$ $$=\lim_{Y \rightarrow \infty} Y \times \left(\frac{\ln x}{Y} - O\left(\frac{\ln X}{Y}\right)^2\right) \ = \ \lim_{Y \rightarrow \infty} \ln x - O\left(\frac{\ln^2 X}{Y}\right)$$ $$= \ln x.$$

The third to last equation due to the Taylor expansion of $\ln y$ for $y$ close to $1$.

Mike
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