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As put forth in the title, the question is to find:

$$\lim_{n\to\infty}(2-\sqrt[n]{2})^n$$

Wolfram Alpha leads me to believe that the correct answer is $\dfrac{1}{2}$.

Thank you!

edit: $n$ is going to infinity. I tried taking the natural log, and did this did not help me. I also attempted to rewrite the term as $$ \left[(1+(1-\sqrt[n]{2}))^{\frac{1}{1-\sqrt[n]{2}}}\right]^{n(1-\sqrt[n]{2})}, $$ but also got nowhere.

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    $\begingroup$ A very important detail which is messing in your question!!! Where is $n$ going? $n \rightarrow ?$. Secondly, why $Wolfram$?... just asking :) $\endgroup$ – Sergio Sarmiento Jun 4 '14 at 13:43
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    $\begingroup$ Add your research effort to the question. $\endgroup$ – Apurv Jun 4 '14 at 13:44
  • $\begingroup$ I tried Wolfram because I thought I got the answer to be 1 at some point and did not believe it. Plugging in high values of $n$ led me to 1/2. I discovered my mistake quickly after realizing I "must" be wrong. $\endgroup$ – user88849 Jun 4 '14 at 13:58
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    $\begingroup$ The form of the problem suggests the definition of $e$. See robjohn's comment for the development. $\endgroup$ – Snowbody Jun 4 '14 at 18:35
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We have

$$\log(2-\sqrt[n]2)=\log(1+(1-\sqrt[n]2))$$ and $$1-\sqrt[n]2=1-\exp\left(\frac1n\log2\right)\sim_\infty-\frac{\log2}n$$ hence we see that $$\log(2-\sqrt[n]2)\sim_\infty -\frac{\log2}n$$ so finally $$\lim_{n\to\infty}(2-\sqrt[n]{2})^n=\lim_{n\to\infty}\exp\left(-\frac{n\log2}n\right)=\frac12$$

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This solution uses L'Hopital's rule. $$\lim_{n\to \infty}(2-\sqrt[n]{2})^n=\lim_{n\to \infty}e^{n\ln (2-\sqrt[n]{2})}=e^{\lim_{n\to \infty}\frac{\ln(2-\sqrt[n]{2})}{1/n}}$$

$$\lim_{n\to \infty}\frac{\ln(2-\sqrt[n]{2})}{1/n}=\lim_{n\to \infty}\frac{\frac{1}{2-\sqrt[n]{2}}(\frac{\ln 2}{n^2}\sqrt[n]{2})}{-1/n^2}=\lim_{n\to \infty}\frac{1}{2-\sqrt[n]{2}}\lim_{n\to \infty}-(\ln 2 )\sqrt[n]{2}=-\ln 2$$

Now plug in to get $e^{-\ln 2}=1/2$

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Note that $$ \begin{align} \left[1+\frac{n\left(2^{1/n}-1\right)}n\right]^n &=\left[2^{1/n}\right]^n\\ &=2\tag{1} \end{align} $$ Since $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag{2} $$ converges uniformly on compact subsets of $\mathbb{R}$ and $e^x$ is strictly increasing, $(1)$ and $(2)$ imply $$ \lim_{n\to\infty}n\left(2^{1/n}-1\right)=\log(2)\tag{3} $$ Again, because $(2)$ converges uniformly on compact subsets of $\mathbb{R}$, $$ \begin{align} \lim_{n\to\infty}\left(2-2^{1/n}\right)^n &=\lim_{n\to\infty}\left(1+1-2^{1/n}\right)^n\\ &=\lim_{n\to\infty}\left[1+\frac{n\left(1-2^{1/n}\right)}n\right]^n\\ &=\lim_{n\to\infty}\left[1+\frac{-\log(2)}n\right]^n\\[9pt] &=e^{-\log(2)}\\[12pt] &=\frac12\tag{4} \end{align} $$

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  • $\begingroup$ Oops, was meant to be a comment on the question pointing out your answer, which seems to me the best way to get the answer $\endgroup$ – Snowbody Jun 4 '14 at 18:36
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Hint: if $a_n>0$ for all $n$ and $\lim_{n\to\infty} a_n = a \neq 0,$ then $\lim_{n\to\infty} \ln(a_n) = \ln a.$

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  • $\begingroup$ $\;a_n, a>0\;$ ........... $\endgroup$ – DonAntonio Jun 4 '14 at 13:53

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