For $ a>0$, one can show, with the help of the AM-GM inequality, that $a + \frac{1}{a} \ge 2 $ and the equality holds at $a=1.$ Therefore, let for some $a>1, a+ \frac{1}{a} = \sqrt{5}.$
$$ \implies x_{2} = \left(a+\frac{1}{a} \right)^2 -2 = a^2 + \frac{1}{a^2}. $$
Similarly, it can be shown that
$$ x_{3} = a^4 + \frac{1}{a^4} $$
Using induction, one can show that
$$ x_{n} = a^{2^{n-1}} + \frac{1}{a^{2^{n-1}}} $$
The given limit becomes
$$ \underset{n \rightarrow \infty}{\lim} \frac{(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4}) \cdots (a^{2^{n-1}} + \frac{1}{a^{2^{n-1}}}) }{a^{2^{n}} + \frac{1}{a^{2^{n}}}} $$
$$ \underset{n \rightarrow \infty}{\lim} \frac{(a- \frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4}) \cdots (a^{2^{n-1}} + \frac{1}{a^{2^{n-1}}}) }{(a-\frac{1}{a})(a^{2^{n}} + \frac{1}{a^{2^{n}}})} $$
$$ \underset{n \rightarrow \infty}{\lim} \frac{(a^{2^{n}} - \frac{1}{a^{2^{n}}})}{(a-\frac{1}{a})(a^{2^{n}} + \frac{1}{a^{2^{n}}})} $$
Since $a>1, $ therefore $ \underset{n \rightarrow \infty}{\lim} a^{2^n} \rightarrow \infty $
$$ \implies \underset{n \rightarrow \infty}{\lim} \frac{(a^{2^{n}} - \frac{1}{a^{2^{n}}})}{(a^{2^{n}} + \frac{1}{a^{2^{n}}})} =1 $$
Therefore, the given limit is equal to:
$$ \frac{1}{a-\frac{1}{a}}. $$
Notice that $a>1 \implies a- \frac{1}{a}>0.$ Also notice that
$$ (a+\frac{1}{a})^2 - (a-\frac{1}{a})^2 = 4 = 5 - (a-\frac{1}{a})^2 $$
$$ \implies (a-\frac{1}{a})^2 =1 \implies a-\frac{1}{a} =1 $$
Which is the value of the given limit.
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