Limits at its best

too Easy Problem ! This is form of ${ 1 }^{ \infty }$ hence I repetedly use formula for this form , I don't give proof since It is useless. $\displaystyle{L=\lim _{ x\rightarrow \infty }{ { \left( f\left( x \right) \right) }^{ { g\left( x \right) } } } ={ e }^{ \lim _{ x\rightarrow \infty }{ (f\left( x \right) -1)g\left( x \right) } }}$

$\displaystyle{\\ L=\lim _{ x\rightarrow \infty }{ { \left( \cos { \left( 2\ln { \left( \cfrac { x-1 }{ x+1 } \right) } \right) } \right) }^{ { (x+1) }^{ 2 } } } \\ L={ e }^{ \lim _{ x\rightarrow \infty }{ { \left( \cos { \left( 2\ln { \left( \cfrac { x-1 }{ x+1 } \right) } \right) } -1 \right) { (x+1) }^{ 2 } } } }\\ L={ e }^{ \lim _{ x\rightarrow \infty }{ { \left( -2\sin ^{ 2 }{ \ln { \left( \cfrac { x-1 }{ x+1 } \right) } } \right) { (x+1) }^{ 2 } } } }\\ L={ e }^{ \left( -2 \right) \lim _{ x\rightarrow \infty }{ { ({ \ln { \left( \cfrac { x-1 }{ x+1 } \right) } ) }^{ 2 }{ (x+1) }^{ 2 } } } }\\ L={ e }^{ \left( -2 \right) \lim _{ x\rightarrow \infty }{ { ((x+1){ \ln { \left( \cfrac { x-1 }{ x+1 } \right) } })^{ 2 } } } }={ e }^{ -2{ { L }_{ 1 } }^{ 2 } }\quad .\quad .\quad .(1)\\ { L }_{ 1 }=\ln { \left( \lim _{ x\rightarrow \infty }{ { (\cfrac { x-1 }{ x+1 } })^{ x+1 } } \right) } \quad ({ 1 }^{ \infty }\quad form)\\ { L }_{ 1 }=\ln { { (e }^{ -2 }) } =-2\quad .\quad .\quad .(2)\\ \boxed { L={ e }^{ -8 } } }$

ans is e^(-8)