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How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?

I ran into this identity on this stackoverflow question. I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure.

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$\lim_{n\to\infty}n(x^{\frac{1}{n}}-1)=\lim_{n\to\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}=f^{\prime}(0)$, where $f(t)=x^t$. Since $$ f^{\prime}(t)=\ln(x)x^t$$ it follows that $f^{\prime}(0)=\ln(x)$.

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    $\begingroup$ Haha this question is worth favoriting just for this elegant proof alone! $\endgroup$ – Chill2Macht May 3 '16 at 5:15
  • $\begingroup$ Given the OP's definition of $\log$, the logic here is circular. $\endgroup$ – yurnero Feb 14 at 11:47
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Set $x=e^t$, then $$ \begin{align} \lim_{n\to\infty}n\left(x^{1/n}-1\right) &=t\lim_{n\to\infty}\frac{e^{t/n}-1}{t/n}\\ &=t\lim_{u\to0}\frac{e^u-1}{u}\\[4pt] &=t\\[8pt] &=\log(x) \end{align} $$

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It depends on how you define $\log x$. In fact this limit itself can be used as a definition of $\log x$ and one can develop full theory of exponential and logarithmic functions starting from this definition.

If we use the definition $$\log x = \int_{1}^{x}\frac{dt}{t}\tag{1}$$ then it is easy to see that $\log x$ is strictly increasing for $x > 0$ and hence possesses an inverse. The inverse function is denoted by $\exp(x)$ or $e^{x}$ and it is defined by equation $$\exp(x) = y \Leftrightarrow x = \log y$$ Using these functions it is possible to prove that $$x^{1/n} = \exp\left(\frac{\log x}{n}\right)$$ Further note that definition $(1)$ implies $$\frac{d}{dx}\log x = \frac{1}{x}$$ and hence derivative of $\log x$ at $x = 1$ is $1$. This means that $$\lim_{h \to 0}\frac{\log(1 + h) - \log 1}{h} = 1$$ or $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1\tag{2}$$ Putting $\log(1 + h) = t$ we see that $$\lim_{t \to 0}\frac{e^{t} - 1}{t} = 1\tag{3}$$ We can now see that \begin{align} L &= \lim_{n \to \infty}n(x^{1/n} - 1)\notag\\ &= \lim_{n \to \infty}n\left(\exp\left(\frac{\log x}{n}\right) - 1\right)\notag\\ &= \lim_{n \to \infty}\log x \cdot \dfrac{\left(\exp\left(\dfrac{\log x}{n}\right) - 1\right)}{\dfrac{\log x}{n}}\notag\\ &= \log x \lim_{t \to 0}\frac{e^{t} - 1}{t}\text{ (putting }t = (\log x)/n)\notag\\ &= \log x\notag \end{align}

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You can even do a bit more using Taylor series $$x^{\frac 1n}=e^{\frac 1 n \log(x)}=1+\frac{\log (x)}{n}+\frac{\log ^2(x)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$n(x^{\frac 1n} -1)=\log (x)+\frac{\log ^2(x)}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

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Starting with the indicated integral, first note that for any $0<a<b$ $$ \int_{a}^b\frac{dt}t=\int_{a}^b\frac{d(at)}{at}=\int_1^{b/a}\frac{dt}t $$ from which the logarithm identity $\ln(b)-\ln(a)=\ln(b/a)$ follows.

Per induction $\ln(a^n)=n\ln(a)$ and thus $$ \ln(x)=n\ln(\sqrt[n]x)=n\int_1^{\sqrt[n]x}\frac{dt}t=n\left(\sqrt[n]x-1\right)\frac1{1+\theta_n(\sqrt[n]x-1)},\quad \theta_n\in(0,1) $$ the last by the mean value theorem in integral form. Thus $$ n\left(\sqrt[n]x-1\right)=\ln(x)·\left(1+\theta_n(\sqrt[n]x-1)\right) $$ and as $\sqrt[n]x\to 1$ for $n\to\infty$, the right side converges to $\ln(x)$.


Note that these calculations avoid any knowledge about properties and derivatives of the exponential function.

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Here is a way to do it with easy integration like you intended to do it. Let $f_n: [a, \infty) \to \mathbb{R}, f_n(t)=t^{-1+\frac{1}{n}}$, with abitrary $a>0$. Because $(f_n)$ converges uniformly $\to t^{-1}$, we can use $$\lim_{n \to \infty} \int_{1}^x f_n (t) dt= \int_{1}^x \lim_{n \to \infty} f_n (t) dt\quad \forall x\in [a, \infty).$$ $$ \iff \lim_{n \to \infty} \int_{1}^x t^{\frac{1}{n}-1} dt=\int_{1}^x \frac{1}{t} dt$$ $$ \iff \lim_{n \to \infty} n(x^{1/n}-1)=\ln{x}\quad \forall x \in [a, \infty).$$ Since $a$ was abitrary $\implies \forall x \in (0, \infty)$.

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This might be helpful if someone is not understanding the above equations.
This is just an expanded version of this equation second answer

$$\lim_{n \to \infty} n (x^{1/n} - 1) = \lim _{n \to \infty} n (e^{ln(x)/n} - 1)$$
then,
$$\lim_{n \to \infty} \frac{e^{ln(x)/n} - 1}{1/n}$$
$$ln(x) \lim_{n \to \infty} \frac{e^{ln(x)/n} - 1}{ln(x)/n}$$
let $\frac{ln(x)}{n}$ be $u$ $$ln(x) \lim_{n \to \infty} \frac{e^{u} - 1}{u}$$
$$ln(x) \lim_{u \to 0} \frac{e^{u} - 1}{u}$$ $$ln(x) \frac{e^{0} - 1}{0}$$ $$ln(x)$$

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  • $\begingroup$ Welcome to MSE. This problem is about $\lim_{n\to\infty}n\left(x^{1/n}-1\right)$, not about $\lim_{x\to\infty}n\left(x^{1/n}-1\right)$. $\endgroup$ – José Carlos Santos Mar 30 at 5:58
  • $\begingroup$ oops, it was a typing mistake. I will correct it. $\endgroup$ – Bipin Maharjan Mar 31 at 6:15

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