Question about formula for natural log

[Matt Benesi]

Any information on the following formula for natural logarithm (I looked in wikipedia and Mathworld but didn't see it). It came from another equation I was working on a bit ago, and I was curious about it as I didn't recall seeing it before (which doesn't mean I haven't), although it reminded me of some equations for e.

$$\ln{x} =\lim_{n\to\infty} \left[ \left (1- \frac{1}{x^{\frac{1}{n}}} \right) \times n \right ]$$
For better visibility (bottom of the fraction is the nth root of x):
$$\ln{x} =\lim_{n\to\infty} \left[ \left (1- \frac{1}{ \sqrt[n]{x}} \right) \times n \right ]$$

Or yet another form:

$$\ln{x} =\lim_{n\to\infty} \left[ \left (1- x^{- \frac{1}{n}} \right) \times n \right ]$$
And I might have answered my own question with this last one... sheesh... anyways, still would like to read about it.
$$\ln{x} =\lim_{n\to\infty} \left[n- n\times x^{- \frac{1}{n}} \right ]$$

Makes the derivative readily apparent, ehh? :D

 

[lurflurf]

It follows from
$$e^x =\lim_{n\to\infty} \left (1+\frac{x}{n}\right)^n$$

 

[Matt Benesi]

Thanks, I realized that post-post, right after I reformulated it a last time and powered down the computer. Ended up writing it out on paper and deriving this particular formula for $e^x$:

$$x = \lim_{n\to\infty} \left(1- \frac{\ln{x}}{n} \right )^{-n}$$
which is basically the following reformulated~~~
$$e^x = \lim_{n\to\infty} \left(1- \frac{x}{n} \right )^{-n}$$