Sparking Recursion

Rearrange the given equation as follows -

$a_{n+1}^2a_n +2a_{n+1}-a_n =0$

$\Rightarrow a_{n+1}=\frac{a_n-a_{n+1}}{1+a_{n+1}a_n}$

The above form gives us a hint of using the following substitution -

$a_n = \tan \theta_n$

Thus,

$a_{n+1} = \frac{\tan \theta_n - \tan \theta_{n+1}}{1+\tan \theta_n \tan \theta_{n+1}} = \tan (\theta_n-\theta_{n+1})$

$\Rightarrow \tan \theta_{n+1} = \tan (\theta_n-\theta_{n+1})$

Now, the principal solution of the above recurrence is

$\theta_{n+1} = \theta_n-\theta_{n+1} \Rightarrow \theta_{n+1} = \frac{\theta_n}{2}$

We don't consider the general solution because we'll end up with the same result.

Notice that, $a_0 = \tan \theta_0 = 1 \Rightarrow \theta_0 = \frac{\pi}{4}$

Therefore, the solution for the recurrence is -

$\theta_n = \frac{\theta_{n-1}}{2} = \frac{\theta_{n-2}}{2^2} = \cdots = \frac{\theta_{0}}{2^n} = \frac{\pi}{2^{n+2}}$

Hence, the final solution of our initial recurrence $a(n)$ is $a_n = \tan \left(\frac{\pi}{2^{n+2}}\right)$

Now, we can easily evaluate the given limit by using the following $\lim_{x \rightarrow \infty} \frac{\tan 1/x}{1/x} = 1$

$\lim_{n \rightarrow \infty} a_n2^n = 2^n\tan \left(\frac{\pi}{2^{n+2}}\right) = \dfrac{\tan \left(\frac{\pi}{2^{n+2}}\right)}{\frac{\pi}{2^{n+2}}} \frac{\pi}{4} = \frac{\pi}{4} = \boxed{0.785}$

First Use Shridhacharya formula for quadratic polynomial , and get : ${ a }_{ n+1 }=\sqrt { 1+{ { a }_{ n } }^{ 2 } } -1$ .

Now think for some substitution ... yes trigonometry rock this time ! Just think..... So I'am only giving outline hints. Solve rest by your own... Clearly

${ a }_{ 0 }=1\quad ,{ a }_{ 1 }=\sqrt { 2 } -1\quad ....$

Now think ..... Yes It is

$\displaystyle{{ a }_{ n }=\tan { \left( \cfrac { { \theta }_{ o } }{ { 2 }^{ n } } \right) } \quad :\quad { \theta }_{ o }=\cfrac { \pi }{ 4 } \\ L=\lim _{ n\rightarrow \infty }{ { 2 }^{ n }\tan { \left( \cfrac { { \theta }_{ o } }{ { 2 }^{ n } } \right) } } ={ \theta }_{ o }=\cfrac { \pi }{ 4 } \\ Ans.}$