Limit of $(2-\sqrt[n]{2})^n$? 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.
 A: 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
$$\lim_{n\to\infty}(2-\sqrt[n]{2})^n=\lim_{n\to\infty}\exp\left(-\frac{n\log2}n\right)=\frac12$$
A: 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 $e^{-\ln 2}=1/2$
A: 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 $\mathbb{R}$,
$$
\begin{align}
\lim_{n\to\infty}\left(2-2^{1/n}\right)^n
&=\lim_{n\to\infty}\left(1+1-2^{1/n}\right)^n\\
&=\lim_{n\to\infty}\left[1+\frac{n\left(1-2^{1/n}\right)}n\right]^n\\
&=\lim_{n\to\infty}\left[1+\frac{-\log(2)}n\right]^n\\[9pt]
&=e^{-\log(2)}\\[12pt]
&=\frac12\tag{4}
\end{align}
$$
A: Hint: if $a_n>0$ for all $n$ and $\lim_{n\to\infty} a_n = a \neq 0,$ then $\lim_{n\to\infty} \ln(a_n) = \ln a.$
