Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a real number $x\geq 1$. Find the limit $$\lim_{n\rightarrow +\infty }\left ( 2\sqrt[n]{x}-1 \right )^n$$

share|improve this question
3  
HINT: use series expansion –  freak_warrior Nov 3 '13 at 1:45

2 Answers 2

$$\lim_{n\to\infty}(2\cdot\sqrt[n]x-1)^n=\lim_{n\to\infty}\exp\,\left[n\cdot\ln\left(2\cdot\sqrt[n]x-1\right)\right]=\lim_{n\to\infty}\exp\left[\frac{\ln(2\cdot x^\frac1n-1)}{\frac1n}\right]=$$ $$=\lim_{n\to\infty}\exp\Bigg[\frac{2\cdot x^\frac1n\cdot\ln x\left(-\frac1{n^2}\right)\bigg/\left(2\cdot\sqrt[n]x-1\right)}{-\frac1{n^2}}\Bigg]=\lim_{n\to\infty}\exp\left[\frac{2\cdot\sqrt[n]x\cdot\ln x}{2\cdot\sqrt[n]x-1}\right]=$$ $$=\exp\left[\frac{2\cdot1\cdot\ln x}{2\cdot1-1}\right]=\exp\left[2\cdot\ln x\right]=x^2$$

share|improve this answer
    
Please tell my why you have $$\lim_{n\to\infty}\exp\left[\frac{\ln(2\cdot x^\frac1n-1)}{\frac1n}\right]=$$$$\lim_{n\to\infty}\exp\Bigg[\frac{2\cdot x^\frac1n\cdot\ln x\left(-\frac1{n^2}\right)\bigg/\left(2\cdot\sqrt[n]x-1\right)}{-\frac1{n^2}}\Bi‌​gg]$$ –  marian porian Nov 3 '13 at 11:43
    
L'Hopital's rule. –  Lucian Nov 3 '13 at 12:43

The Taylor expansion of $x^{n}$ at $n=0$ is $\displaystyle 1+ n \ln x + \mathcal{O}(n^{2})$.

So the Laurent expansion of $x^{\frac{1}{n}}$ at $n= \infty$ is $ \displaystyle 1 + \frac{\ln x}{n} + \mathcal{O}\left( \frac{1}{n^{2}}\right)$.

Then $ \displaystyle \lim_{n\to \infty }\left( 2\sqrt[n]{x}-1 \right )^{n} = \lim_{n \to \infty} \left(2 + \frac{2 \log x}{n} + \mathcal{O} (n^{-2} )-1 \right)^{n} $

$\displaystyle = \lim_{n \to \infty} \left( 1 + \frac{\log x^{2}}{n} + \mathcal{O}(n^{-2}) \right)^{n} = e^{\log x^{2}} = x^{2}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.