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Prove that $\lim_{n\to 0}\frac {x^n-1}{n} = \ln x$

This expression appears while calculating the derivative of exponential function

Mark Viola
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Zyad Yasser
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    What is your definition of the logarithm? It's impossible to answer this consistently without that. – Chappers Jul 26 '17 at 20:26
  • With $x=e^t$ and $m=nt$ and $t\ne 0$,

    $$\lim_{n\to0}\frac{x^n-1}n=\lim_{n\to0}\frac{e^{nt}-1}n=t\lim_{m\to0}\frac{e^{m}-1}m=ct=c\ln x$$ where $c$ is some constant, if the limit exists.

     –  Jul 26 '17 at 20:36
  • This question itself is good and if kept open would doubtless get many answers that are merely correct (and "merely" is the right word for that), and possibly also one or more good answers. It was closed because the poster phrased it as if the poster were assigning us homework. Is there any possibility of closing it as a duplicate instead? – Michael Hardy Jul 26 '17 at 20:40
  • @ZezoYasor : Since you say where this appeared, if you post some specifics about that, then there's a good chance this question can be re-opened. – Michael Hardy Jul 26 '17 at 20:50
  • Indeed, I concur with @MichaelHardy. I will certainly vote to reopen, Zezo, you're willing to provide more details about the question from which your question emerges. – amWhy Jul 26 '17 at 21:36

1 Answers1

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HINT:

In This Answer, I used the limit definition of the exponential function, $e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$, along with Bernoulli's Inequality to arrive at the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag1$$

for $x<1$.

Then defining $\log(x)$ as the inverse function of $e^x$, note that

$$\frac{x^n-1}{n}=\frac{e^{n\log(x)}-1}{n}\tag 2$$

Finally, apply $(1)$ to $(2)$.

Mark Viola
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