TITLE: Limit of $(2-\sqrt[n]{2})^n$?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $\mathbb{R}$,
$$
\begin{align}
\lim_{n\to\infty}\left(2-2^{1/n}\right)^n
&=\lim_{n\to\infty}\left(1+1-2^{1/n}\right)^n\\
&=\lim_{n\to\infty}\left[1+\frac{n\left(1-2^{1/n}\right)}n\right]^n\\
&=\lim_{n\to\infty}\left[1+\frac{-\log(2)}n\right]^n\\[9pt]
&=e^{-\log(2)}\\[12pt]
&=\frac12\tag{4}
\end{align}
$$