What is limit of $ \lim \limits_{n \to \infty} \sqrt[n]{\sqrt[2^n]{a}-1}$ given $a>1$? What is the limit of $\displaystyle{ \lim \limits_{n \to \infty} \sqrt[n]{\sqrt[2^n]{a}-1}}$ given $a>1$ ?
I did some computations and I feel its $\frac{1}{2}$ , i don't know how to prove it
I did used Bernoulli inequality 
 A: Let's do this without logarithms, using only some algebra, the limit $\sqrt[n]x\to1$ for any $x\gt0$, and the Squeeze Theorem.
If $A\gt1$ and $N\in\mathbb{N}$, it's easy to see that
$$(A^N-1)=(A-1)(A^{N-1}+\cdots+1)\le(A-1)(A^N+\cdots+A^N)=NA^N(A-1)$$
and
$$A^N-1=(1+(A-1))^N-1=1+N(A-1)+\cdots+(A-1)^N-1\ge N(A-1)$$
hence
$${A^N-1\over NA^N}\le A-1\le{(A^N-1)\over N}$$
Now let $N=2^n$ and $A=a^{1/N}=a^{1/2^n}$ with $a\gt1$.  Then 
$${a-1\over2^na}\le a^{1/2^n}-1\le{a-1\over2^n}$$
and so
$${1\over2}\sqrt[n]{a-1\over a}\le\sqrt[n]{a^{1/2^n}-1}\le{1\over2}\sqrt[n]{a-1}$$
Since $a-1$ and $(a-1)/a$ are positive, the upper and lower bounds tend to $1/2$ as $n\to\infty$, so the Squeeze Theorem tells us
$$\lim_{n\to\infty}\sqrt[n]{a^{1/2^n}-1}={1\over2}$$
A: Note that$$\lim_{n\to\infty}n\left(\sqrt[n]a-1\right)=\log a\tag{1}.$$This equality comes from $$\lim_{n\to\infty}n\left(\sqrt[n]a-1\right)=\lim_{n\to\infty}\frac{a^\frac1n-1}{\frac1n},$$which is the derivative at $0$ of the function $t\mapsto a^t$. Since $a^t=\exp\bigl(t\log(a)\bigr)$, this derivative is $\log a$.
From $(1)$, you can deduce that$$\lim_{n\to\infty}2^n\left(\sqrt[2^n]a-1\right)=\log a\in(0,+\infty).$$But then$$\lim_{n\to\infty}\sqrt[n]{2^n\left(\sqrt[2^n]a-1\right)}=1,$$which means that$$\lim_{n\to\infty}\sqrt[n]{\sqrt[2^n]a-1}=\frac12.$$
A: The limit in question is reduced to the limit
$$
\lim_{t \to 0_+} (a^t-1)^{1/\ln(t)}=e
$$
which is worth remembering. 
