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Let $f_n(x) = x^{1/n}$ where $n \in \mathbb N$, and let $g(x) = \log(x)$. We can compute $f_n'(x) = \frac{1}{n}x^{-1 + \frac{1}{n}}$ and $g'(x) = x^{-1}$. Let's define $f_\infty(x) = \lim_{n \rightarrow \infty} f_n(x)$. It will be everywhere assumed that $x > 0$.

If I've computed it correctly, it seems that $\lim_{n \rightarrow \infty} f_n'(x) = 0$ for any fixed $x > 0$. This means that for each $x \in \mathbb R^+$ $f_\infty$ is flat while $g$ is not.

But if instead we compute $\lim_{x \rightarrow \infty} \frac{f_n(x)}{g(x)}$ for a fixed $n \in \mathbb N$ then we find that this limit equals $\infty$, suggesting that for each finite $n$ $f_n$ grows faster.

Initially I was thinking that maybe $\log$ was like the limiting case of $f_n$ but now I'm not so sure. Part of the issue is reconciling these two limits.

My question, then, is: can we think of $\log$ as the limiting case of $f_n$? If not, why?

alfalfa
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    Observe that $;\displaystyle\int x^{-1+1/n} , dx=\frac{x^{1/n}}{1/n} + C=\frac{x^{1/n}-1}{1/n} +C';$ and that the limit as $n\to\infty$ will indeed be $\log(x)$ from this thread. – Raymond Manzoni Apr 13 '16 at 15:29
  • @RaymondManzoni that is very helpful. If you post that as an answer I'll accept it. – alfalfa Apr 13 '16 at 16:33
  • Glad it clarified things @alfalfa! Fine continuation, – Raymond Manzoni Apr 13 '16 at 21:44
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    It is not difficult to develop a theory of logarithmic and exponential functions starting from the definition $\log x = \lim\limits_{n \to \infty}n(x^{1/n} - 1)$. See http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-2_10.html – Paramanand Singh Apr 14 '16 at 09:23
  • @ParamanandSingh Thanks for the link. Now that you point that out, it seems really obvious that $(1 + \frac{x}{n})^n = y \implies n(y^{1/n} - 1) = x$ so this clearly is the natural choice – alfalfa Apr 15 '16 at 14:08

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Short answer : $\;f_n:=n\,\left(x^{1/n}-1\right)\;$ will give $\,\log(x)$ as $\,n\to\infty$ rather than $\,f_n:=x^{1/n}$.

Observe that for any positive $n$ we have : $$\int x^{-1+1/n} \, dx=\frac{x^{1/n}}{1/n} + C=\frac{x^{1/n}-1}{1/n} +C'=\frac{e^{\,\large{\log(x)/n}}-1}{1/n} +C'$$ while the limit as $n\to\infty$ of the fraction obtained will be (for any fixed positive $x$) : $$\lim_{n\to +\infty} \frac{e^{\,\large{\log(x)/n}}-1}{1/n}=\lim_{n\to +\infty} \frac{\log(x)/n}{1/n}=\log(x) $$ For alternative answers see this thread.

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Raymond Manzoni
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