If a sequence $langle a_nrangle$ is such that $a_1a_2=1, a_2a_3=2, ldots$ and $limfrac{a_n}{a_{n+1}}=1.$ Then find $|a_1|.$

Note that $a_{2k+1} = frac{2k}{2k-1}a_{2k-1} = ... = a_1prod_{j=1}^k frac{2j}{2j-1} $ and $a_{2k+2} = a_2 prod_{j=1}^k frac{2j+1}{2j} = frac{1}{a_1}prod_{j=1}^k frac{2j+1}{2j}$

By looking at the subsequence: $$ 1 = lim_{k to infty} frac{a_{2k+1}}{a_{2k+2}} = a_1^2lim_{k to infty} frac{prod_{j=1}^k frac{2j}{2j-1}}{prod_{j=1}^k frac{2j+1}{2j}} = a_1^2 lim_{k to infty} prod_{j=1}^k frac{(2j)(2j)}{(2j+1)(2j-1)} = a_1^2 frac{pi}{2} $$ Where we used Wallis Formula for $pi$: $$ prod_{j=1}^infty frac{4j^2}{(2j-1)(2j+1)} = frac{pi}{2} $$

Similarly $$ 1 = lim_{k to infty} frac{a_{2k+2}}{a_{2k+3}} = frac{1}{a_1^2} lim_{k to infty} frac{2k+1}{2k+2} cdotprod_{j=1}^k frac{(2j+1)(2j-1)}{(2j)(2j)} = frac{1}{a_1^2}frac{2}{pi}$$

Hence in both cases (and subsequences $(2k)_{k in mathbb N},(2k+1)_{k in mathbb N}$ cover whole sequence $(n)_{n in mathbb N}$), we get condition $$a_1^2 = frac{2}{pi}$$ so we get the value of $a_1$ up to the sign, that is: $$ |a_1| = sqrt{frac{2}{pi}}$$

And Mike F already showed that $|a_1|$ is unique

You can easily get a uniqueness statement: suppose that $a_n$ and $b_n$ are two sequences satisfying the givens. Notice that $a_n$ and $b_n$ are never zero, because of the condition on products. Let $c_n = frac{a_n}{b_n}$. Then we have $c_nc_{n+1} = 1$ and $lim frac{c_n}{c_{n+1}}=1$. If we let $c_1=x$, the first condition clearly specifies the whole sequence as

begin{align*} c_1=x && c_2=frac{1}{x} && c_3 = x && c_4 = frac{1}{x} && ldots end{align*}

The condition on the ratios is then that

begin{align*} x^2, frac{1}{x^2}, x^2, frac{1}{x}^2, ldots to 1 end{align*}

which gives $x= pm 1$. So, either $a_n = b_n$ for all $n$, or else $a_n = -b_n$ for all $n$. Thus the solution, if it exists, is unique up to multiplying the whole sequence by $-1$ (which clearly preserves the given conditions).