<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] 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.
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