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

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HINT: use series expansion – freak_warrior Nov 3 '13 at 1:45

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

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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} $

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

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