Compute $\lim_\limits{n\to\infty}a_n$ where $a_{n+2}=\sqrt{a_n.a_{n+1}}$ I managed to show that the limit exists, but I don't know how to compute it.
EDIT:
There are initial terms: $a_1=1$ and $a_2=2$.
 A: Note that $$  a_{n+2}\sqrt{a_{n+1}}=a_{n+1}\sqrt{a_n} =\cdots =a_2\sqrt{a_1}=2$$
Hence limit is $2^\frac{2}{3}$. 
A: With $b_n = \log a_n$ (as suggested in a comment) the iteration formula
becomes
$$
  b_{n+2} = \frac 12 (b_n + b_{n+1}) \, .
$$
Computing the first iterates numerically leads to the conjecture
that the sequence converges to $(b_0 + 2b_1)/3\, $.
Therefore we define
$$
  b := \frac{b_0 + 2b_1}3 \, , \quad c := \frac{b_1 - b_0}3 \, .
$$
Then
$$
  b_0 = b - 2c \,, \quad b_1 = b + c \\
  b_2 = b - \frac c2\,, \quad b_3 = b + \frac c4 \\
  b_4 = b - \frac c8\,, \quad b_5 = b + \frac c{16} \\
$$
and generally
$$
b_n = b + \frac{(-1)^{n-1}c}{2^{n-1}}
$$
It follows that
$$
  \lim_{n \to \infty} b_n = b = \frac{b_0 + 2b_1}3
$$
and therefore
$$
  \lim_{n \to \infty} a_n = \sqrt[3]{a_0 a_1^2}
$$
In your case ($a_0 = 1$, $a_1 = 2$) the limit is $2^{2/3} \approx 1.587401$.
A: Looking at the sequence, we first see that if $a_n' = \alpha a_n$ and $a_{n+1}' = \alpha a_{n+1}$, then also $a_{n+2}' = \alpha a_{n+2}$, and therefore also for the limit $a := \lim_{n\to\infty}a_n$ we have $a' = \alpha a$. Therefore we can write
$$a = a(a_1, a_2) = a_1 a(1, a_2/a_1) =: a_1 f(a_2/a_1).$$
Now what properties does $f$ have? Well, we can of course start the sequence at every position, therefore we have
$$f(x) = a(1, x) = a(x, \sqrt{x}) = x a(1, x^{-1/2}) = xf(x^{-1/2}).$$
Also we can see that if $a_n' = a_n^\alpha$ and $a_{n+1}' = a_{n+1}^{\alpha}$, then also $a_{n+2}' = a_{n+2}^\alpha$, and thus
$$f(x^\alpha) = f(x)^\alpha.$$
Thus we have
$$f(x) = xf(x^{-1/2}) = x(f(x))^{-1/2}$$
and therefore
$$f(x)^{3/2} = x \implies f(x) = x^{2/3}$$
Thus we have
$$a = a_1 f(a_2/a_1) = a_1 \left(\frac{a_2}{a_1}\right)^{\frac23} = a_1^{1/3} a_2^{2/3}.$$
In particular, with $a_1=1$ and $a_2=2$ we get
$$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$
A: We have $a_2\gt a_1\gt0$ and $a_{n+2}^2=a_{n+1}a_n$.
So,$$\begin{align}
&a_3^2=a_2a_1\\
\implies&a_3^2\gt a_1^2\\
\implies &a_3\gt a_1
\end{align}$$
and $$\begin{align}
&a_3^2=a_2a_1\\
\implies &a_3^2\lt a_2^2\\
\implies &a_3\lt a_2
\end{align}$$
So, we have $$\color{red}{a_2\gt a_3\gt a_1\gt 0}$$
Similarly, $$\begin{align}
&a_4^2=a_3a_2\\
\implies &a_4^2\lt a_2^2\\
\implies &a_4\lt a_2
\end{align}$$
 and  $$\begin{align}
&a_4^2=a_3a_2\\
\implies &a_4^2\gt a_3^2\\
\implies &a_4\gt a_3
\end{align}$$
So, we have $$\color{red}{a_2\gt a_4\gt a_3\gt a_1\gt 0}$$
Now, following this pattern, one can easily see that this sequence is bounded below by $0$ and bounded above by $a_2$.
Now, consider two subsequences $\color{blue}{v_n=a_{2n-1}}$ and $\color{blue}{w_n=a_{2n}}$.
one can easily see that $v_n$ is increasing subsequence bounded above by $a_2$, since, $\color{blue}{0\lt a_1\lt a_3\lt \dots \lt a_2}$, and $w_n$ is decreasing subsequebce and bounded below by $0$, since, $\color{blue}{a_2\gt a_4\gt \dots \gt 0}$.
So, they must converge. Suppose $\color{blue}{\lim v_n=l_1}$ and $\color{blue}{\lim w_n=l_2}$.
Now, we have, $$\begin{align}
&a_{n+2}^2=a_{n+1}a_n\\
\implies &\lim a_{n+2}^2=\lim a_{n+1}\times\lim a_n\\
\implies &l_2^2=l_1\times l_2\\
\implies &\color{blue}{l_2=l_1=l(\text{say})}
\end{align}$$
here, without loss of generality, we assumed $n$ is even. So, the mother sequence $\{a_n\}$ converges to $l$ i.e. $\color{blue}{\lim a_n=l}$
Now write $$\begin{align}
\color{green}{\cancel{a_3^2}}&=a_2\times a_1\\
\color{purple}{\cancel{a_4^2}}&=\color{green}{\cancel{a_3}}\times a_2\\
\color{orange}{\cancel{a_5^2}}&=\color{purple}{\cancel{a_4}}\times \color{green}{\require{cancel}\cancel{a_3}}\\
&\vdots\\
\color{navy}{a_{n+1}^\cancel{2}}&=\color{maroon}{\cancel{a_{n}}}\times \color{teal}{\cancel{a_{n-1}}}\\
a_{n+2}^2&=\color{navy}{\cancel{a_{n+1}}}\times \color{maroon}{\cancel{a_n}}.
\end{align}$$
Now, multiplying all the terms, we have $$\begin{align}
&a_{n+2}^2 \times a_{n+1}=a_2^2 a_1\\
\implies &\lim a_{n+2}^2 \times \lim a_{n+1}=a_2^2 a_1\\
\implies &l^2\times l= a_2^2 a_1\\
\implies &l^3=a_2^2 a_1\\
\implies &\color{red}{l=\sqrt[3]{a_2^2 a_1}}.
\end{align}$$ 
And, here the particular case, $a_1=1$ and $a_2=2$, so,
$$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$
