Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$ 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.
 A: Set $x=e^t$, then
$$
\begin{align}
\lim_{n\to\infty}n\left(x^{1/n}-1\right)
&=t\lim_{n\to\infty}\frac{e^{t/n}-1}{t/n}\\
&=t\lim_{u\to0}\frac{e^u-1}{u}\\[4pt]
&=t\\[8pt]
&=\log(x)
\end{align}
$$
A: 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}
A: 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.
A: 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.
A: $\lim_{n\to\infty}n(x^{\frac{1}{n}}-1)=\lim_{n\to\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}=f^{\prime}(0)$, where $f(t)=x^t$. Since
$$ f^{\prime}(t)=\ln(x)x^t$$
it follows that $f^{\prime}(0)=\ln(x)$.
A: 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 $\implies \forall x \in (0, \infty)$.
A: 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 $u$
$$\ln(x) \lim_{n \to \infty} \frac{e^{u} - 1}{u}$$
$$\ln(x) \lim_{u \to 0} \frac{e^{u} - 1}{u}$$
$$\ln(x) \frac{e^{0} - 1}{0}$$
$$\ln(x)$$
A: This is mainly to give an elementary proof that makes use of the integral definition of $\ln x$.
Taking for granted the (easily proved) limit $x^{1/n}\to1$ for all $x\gt0$, it suffices to prove the equality between $\ln x$ and $\lim_{n\to\infty}n(x^{1/n}-1)$ for $x\ge1$, since we then have
$$n((1/x)^{1/n}-1)={-n(x^{1/n}-1)\over x^{1/n}}\to{-\ln(x)\over1}=\ln(1/x)$$
Now let's assume that the limit of $n(x^{1/n}-1)$ exists, i.e., there is some function $L(x)=\lim_{n\to\infty}n(x^{1/n}-1)$. (We'll return to this at the end and give a proof of existence.) Then
$$\begin{align}
|n(x^{1/n}-1)-\ln x|
&=|n(x^{1/n}-1)-n\ln(x^{1/n})|\\
&=\left|n\int_1^{x^{1/n}}dt-n\int_1^{x^{1/n}}{dt\over t}\right|\\
&=n\int_1^{x^{1/n}}{t-1\over t}\,dt\\
&\le n\cdot{x^{1/n}-1\over 1}\int_1^{x^{1/n}}dt\\
&=n(x^{1/n}-1)(x^{1/n}-1)\to L(x)(1-1)=0
\end{align}$$
so the squeeze theorem gives the desired equality.
To show that $\lim_{n\to\infty}n(x^{1/n}-1)$ exists, it suffices to show that the sequence $n(x^{1/n}-1)$, which is clearly positive (for $x\ge1$), is monotonically decreasing.  For this, let $u=x^{1/n(n+1)}$, and note that
$$\begin{align}
u^{n-1}+u^{n-2}+\cdots+u+1\le nu^n
&\implies u^n-1\le nu^n(u-1)\\
&\implies(n+1)(u^n-1)\le n(u^{n+1}-1)\\
&\implies(n+1)(x^{1/(n+1)}-1)\le n(x^{1/n}-1)
\end{align}$$
where the opening inequality is true courtesy of the fact that $u^k\le u^n$ for all $k\le n$ when $u\ge1$.
Remark: An easy proof of $x^{1/n}\to1$ follows similar lines: $1\le x^{1/(n+1)}\le x^{1/n}$, so a limit exists, and can only be $1$ by the squeeze theorem, from
$$x^{1/n}-1={x-1\over x^{(n-1)/n}+x^{(n-2)/n}+\cdots+1}\le{x-1\over n}\to0$$
