How to prove this limit: $\lim_{n\to \infty}n(\sqrt[n]{x}-1)=\ln(x)$?

Question

I don't know how to prove that $\forall x\in \mathbb{R^+} \displaystyle \lim_{n\to \infty}n(\sqrt[n]{x}-1)=\ln(x)$

with elementary techniques

Possible duplicate of: http://math.stackexchange.com/questions/1769256/trying-to-show-that-lnx-lim-n-to-infty-nx1-n-1/1769257#1769257 — carmichael561, May 28 '16 at 18:35

score 6 · Accepted Answer · answered May 28 '16 at 18:23

6

Write $$ n(\sqrt[n]{x}-1)=\frac{e^{\frac{1}{n}\ln x}-1}{\frac{1}{n}} $$ and use the notable limit $$ \lim_{t\to 0}\frac{e^t-1}{t}=1\quad (\textrm{which implies }\lim_{t\to 0}\frac{e^{at}-1}{t}=a). $$

answered May 28 '16 at 18:23

guestDiego

4,006

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

2

I thought it would be instructive to see a way forward that forgoes use of calculus. To that end, we proceed.

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}}\tag 1$$

for $x<1$. Then, using $(1)$ along with the identity $\sqrt[n]{x}=e^{\frac1n \log(x)}$, we have

$$\log(x) \le n\left(e^{\frac1n \log(x)}-1\right)\le \frac{\log(x)}{1-\frac1n \log(x)} \tag 2$$

Applying the squeeze theorem to $(2)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}n\left(\sqrt[n]{x}-1\right)=\log(x)}$$

edited Apr 13 '17 at 12:21

Community

1

answered May 28 '16 at 20:36

Mark Viola

179,405

I cant see why $$\left(e^{\frac1n \log(x)}-1\right)\le \frac{\log(x)}{1-\frac1n \log(x)}$$ – Nah May 29 '16 at 00:10
1

From (1), you get $e^u-1 \leq \frac u{1-u}$. Replace $u $ by $\frac{\log x}{n} $, then multiply both sides by $n $. – Clement C. May 29 '16 at 07:15
@clementc. Thanks for the comment. +1 – Mark Viola May 29 '16 at 21:37

How to prove this limit: $\lim_{n\to \infty}n(\sqrt[n]{x}-1)=\ln(x)$?

2 Answers2