Limit of recursive sequence defined by $a_0=0$, $a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right)$ Given the following sequence: $a_0=0$, $$a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right),\ \forall n\ge 0.$$ 
Find $\lim\limits_{n\to\infty}a_n$.

Hint:Take $a_n = \dfrac{\tan \theta_n}{2^n}$
