<s>[INST] Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$ 
Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.

Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find the equation for the $x_n$ term in terms of $3$. The second part can be answered as follows, please tell me if my thinking is correct.
\begin{align*}
x_n&=3^{1-(\frac{1}{2^{n-1}})} \\
\lim_{n\rightarrow \infty}x_n&=\lim_{n\rightarrow \infty}3^{1-(\frac{1}{2^{n-1}})} \\
&=3^{1-\lim_{n\rightarrow \infty}(\frac{1}{2^{n-1}})} \\
&=3^{1-(\frac{1}{\lim_{n\rightarrow \infty}2^{n-1}})} \\
&= 3
\end{align*}
Also I have looked at Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit. But there is no discussion on how to find something similar to $x_n=3^{1-(\frac{1}{2^{n-1}})}$ .
 [/INST] If you want to see where the formula comes from, notice the pattern:
\begin{eqnarray}
x_1 &=& 1\\
x_2 &=& \sqrt{3} &=& 3^{\frac12}\\
x_3 &=& \sqrt{3\sqrt{3}} &=& \bigl(3\cdot3^{\frac12}\bigr)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 &=& 3^{\frac12 + \frac14}\\
x_4 &=& \sqrt{3\sqrt{3\sqrt{3}}} &=& \left(3\bigl(3\cdot3^{\frac12}\bigr)^\frac12\right)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 \cdot 3^\frac18 &=& 3^{\frac12 + \frac14 + \frac18}\\
\end{eqnarray}
In general, the sequence can be written 
$$x_n = 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}$$
Inductively,
\begin{eqnarray}
x_{n+1} &=& \sqrt{3x_n}\\
        &=& \sqrt{3 \cdot 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\
        &=& \sqrt{3^{1+\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\
        &=& 3^{\frac12+\frac12\sum_{k=1}^{n-1}\frac{1}{2^k}}\\
        &=& 3^{\sum_{k=1}^{(n-1)+1}\frac{1}{2^k}}\\
\end{eqnarray}
The sum for the geometric series is $\displaystyle\sum_{k=0}^{n-1} a r^k = a \dfrac{1-r^n}{1-r}$, so
$$1+\sum\limits_{k=1}^{n-1}\frac{1}{2^k} = \sum\limits_{k=0}^{n-1}\frac{1}{2^k} = \frac{1-(\frac12)^n}{1-\frac12} = \frac{1-\frac{1}{2^n}}{\frac12} = 2 - \frac{1}{2^{n-1}}$$
Subtracting $1$ from both sides of the equation, we can write
$$x_n = 3^{\left(1 - \frac{1}{2^{n-1}}\right)}= 3^{1-2^{1-n}}$$
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