How to prove this form of $\ln(x)$

Question

It is well known that $$\lim_{h\to0}\frac{x^h-1}{h}=\ln x$$

but, how can I prove this? You can't use l'hopital because the $\mathrm d/\mathrm dx$ of $x^h$ depends on this limit.

Answer 1

If you want to prove that $\;\lim\limits_{h\to0}\dfrac{x^h-1}h=\ln x\;$ without using the notable limit $\;\lim\limits_{t\to0}\dfrac{e^t-1}t=1\,,\,$ which is very similar to what you want to get, you could proceed in the following way.

Let $\;t=\dfrac1{x^h-1}\;.$

It results that $\;t\to\pm\infty\;$ as $\;h\to0^{\pm}\;,\;$ moreover ,

$h=\dfrac{\ln\left(1+\frac1t\right)}{\ln x}\;,\;$ hence ,

$\lim\limits_{h\to0^{\pm}}\dfrac{x^h-1}h=\lim\limits_{t\to\pm\infty}\dfrac{\frac1t}{\frac{\ln\left(1+\frac1t\right)}{\ln x}}=\lim\limits_{t\to\pm\infty}\dfrac{\ln x}{\ln\left(1+\frac1t\right)^t}=\ln x\;,$

indeed $\;\lim\limits_{t\to\pm\infty}\left(1+\dfrac1t\right)^t=e\;.$

Answer 2

Since

$\displaystyle \lim_{t\to 0}\dfrac{\text{e}^t-1}{t}=1$

It follows

$\displaystyle \lim_{h\to 0}\dfrac{x^h-1}{h}=\lim_{h\to 0}\dfrac{\text{e}^{h\log x}-1}{h}=\lim_{h\to 0}\dfrac{\text{e}^{h\log x}-1}{h\log x}\log x=\log x$