Limiting Trigo 2

@Murugesh M here is the solution please refer to it.

$Let\quad \quad \quad \quad y\quad =\sqrt { (tan\quad x-sin\quad x)+\sqrt { (tan\quad x-sin\quad x)+...\infty } }$

$\Rightarrow \quad \quad \quad \quad y =\sqrt { (tan\quad x-sin\quad x)+y }$

$\Rightarrow \quad \quad \quad \quad { y }^{ 2 }-y-(tan\quad x-sin\quad x)=0$

$\Rightarrow \quad \quad \quad { y=\frac { 1+\sqrt { 1+\left( tan\left( x \right) -sin\left( x \right) \right) 4 } }{ 2 } \quad \quad \quad \quad \left[ \because \quad y>0 \right] ...(i) }$

$Also\quad let,\quad \quad \quad z=\sqrt { { x }^{ 3 }+\sqrt { { x }^{ 3 }+\sqrt { { x }^{ 3 }+...\infty } } }$

$\Rightarrow \quad \quad \quad z=\sqrt { { x }^{ 3 }+z } \quad \quad \Rightarrow { z }^{ 2 }-z-{ x }^{ 3 }=0\quad$

$\Rightarrow \quad \quad z=\frac { 1+\sqrt { 1+4{ x }^{ 3 } } }{ 2 } \quad \quad \quad \left[ \because \quad \quad z>0 \right] ...(ii)\quad$

$\therefore \quad \quad$ From (i) and (ii),

$\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { -1+\sqrt { (tan\quad x-sin\quad x)+\sqrt { (tan\quad x-sin\quad x)+...\infty } } }{ -1+\sqrt { { x }^{ 3 }+\sqrt { { x }^{ 3 }+\sqrt { { x }^{ 3 }+...\infty } } } }$

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { -1+\frac { 1+\sqrt { 1+\left( tan\left( x \right) -sin\left( x \right) \right) 4 } }{ 2 } }{ -1+\frac { 1+\sqrt { 1+4{ x }^{ 3 } } }{ 2 } }$

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { -1+\sqrt { 1+\left( tan\left( x \right) -sin\left( x \right) \right) 4 } }{ -1+\sqrt { 1+4{ x }^{ 3 } } }$

Rationalising numerator and denominator we get,

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { 4\left( tan\left( x \right) -sin\left( x \right) \right) \left( 1+\sqrt { 1+4{ x }^{ 3 } } \right) }{ 4{ x }^{ 3 }\left( 1+\sqrt { 1+4\left( tan\left( x \right) -sin\left( x \right) \right) } \right) }$

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { 4\left( \frac { sin\left( x \right) }{ cos\left( x \right) } -\frac { sin\left( x \right) }{ 1 } \right) \left( 1+\sqrt { 1+4{ x }^{ 3 } } \right) }{ 4{ x }^{ 3 }\left( 1+\sqrt { 1+4\left( tan\left( x \right) -sin\left( x \right) \right) } \right) }$

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { sin\left( x \right) \left( 1-cos\left( x \right) \right) }{ { x }^{ 3 }cos\left( x \right) } .\frac { \left( 1+\sqrt { 1+4{ x }^{ 3 } } \right) }{ \left( 1+\sqrt { 1+4\left( tan\left( x \right) -sin\left( x \right) \right) } \right) }$

$=\begin{matrix} lim \\ x\rightarrow { 0 }^{ + } \end{matrix}\frac { sin\left( x \right) }{ x } .\frac { 2{ sin }^{ 2 }\left( \frac { x }{ 2 } \right) }{ \frac { 4{ x }^{ 2 } }{ 4 } } .\frac { 1 }{ cos\left( x \right) } .\frac { \left( 1+\sqrt { 1+4{ x }^{ 3 } } \right) }{ \left( 1+\sqrt { 1+4\left( tan\left( x \right) -sin\left( x \right) \right) } \right) }$

$=1.\frac { 1 }{ 2 } .1.\frac { \left( 1+1 \right) }{ \left( 1+1 \right) } =\frac { 1 }{ 2 } =\quad \boxed{0.5}$