Cut it the right way

By closely observing the recurrence we see we have to use a substitution :

${T}_{n} = 2cos\left(\dfrac{\theta}{2^{n}}\right)$ , where $\theta$ is an arbitary constant.

You can verify that it satisfies the recurrence relation.

For finding $\theta$ put , $n=0$

$2cos(\theta) = 0$

$\Rightarrow \theta = \dfrac{\pi}{2}$

Hence we got ${T}_{n} = 2cos\left(\dfrac{\pi}{2^{n+1}}\right)$

Hence our limit becomes :

$L =\displaystyle \lim _{ n\rightarrow \infty }{ { 4 }^{ n }\left(3-\sqrt { 3(1+2\cos { \dfrac { \pi }{ { 2 }^{ n+1 } } ) } } \right) }$

Put $\dfrac { \pi }{ { 2 }^{ n+1 } } =x$ to get the limit as :

$L = \displaystyle \dfrac { { \pi }^{ 2 } }{ 4 } \lim _{ x\rightarrow 0 }{ \left(\dfrac { 3-\sqrt { 3(1+2\cos { x) } } }{ { x }^{ 2 } } \right) }$