Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$

Solution 1:

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

Solution 2:

Set $x=e^t$, then $$ \begin{align} \lim_{n\to\infty}n\left(x^{1/n}-1\right) &=t\lim_{n\to\infty}\frac{e^{t/n}-1}{t/n}\\ &=t\lim_{u\to0}\frac{e^u-1}{u}\\[4pt] &=t\\[8pt] &=\log(x) \end{align} $$

Solution 3:

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}