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 $x_n=3^{1-(\frac{1}{2^{n-1}})}$ .
 A: Your proof for the second part is correct.
What's nice here is you don't need to find the equation for $x_n$, it's already given to you. You just have to prove the equation is correct. You can do this by induction on $n$. The base case $n=1$ holds since $x_1=1=3^{1-\frac1{2^{1-1}}}$.
Assume that $x_n=3^{1-\frac1{2^{n-1}}}$. Then
$$
x_{n+1}=\sqrt{3x_n}=\sqrt{3\cdot 3^{1-\frac1{2^{n-1}}}}=3^{1-\frac1{2^{(n+1)-1}}}
$$
This proves the formula for $n+1$, completing the proof by induction.

One last comment: in general, there is no procedure for getting a formula for $x_n$ given a recursive equation is solves. In this case, it could be done by looking at $x_n$ for several small values of $n$, simplifying as much as possible, then noticing a pattern. 
A: 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
$$x_n = 3^{\left(1 - \frac{1}{2^{n-1}}\right)}= 3^{1-2^{1-n}}$$
A: Hint
Considering the sequence $$x_{n+1}=\sqrt{3x_n}$$ take logarithms of both sides $$\log(x_{n+1})=\frac{\log(3)}2+\frac 12 \log(x_n)$$ Now define $y_n=\log(x_n)$ which makes the new sequence to be $$y_{n+1}=\frac 12 y_n+\frac{\log(3)}2$$ which is much easier to manipulate.
For the general sequence $$y_{n+1}=a y_n+b\qquad (y_1=0)$$ the general solution is simply $$y_n=\frac{1-a^{n-1}}{1-a} b$$ So, if $a<1$, the limit is just $\frac b {1-a}$; then the limit for the original  sequence.
