Limiting Trigo...

Let $y\quad =\quad \sqrt { \alpha +\sqrt { \alpha +\sqrt { \alpha +......\infty } } }$

where $\alpha = 1 - cos \quad x$ ; $as \quad x\rightarrow 0, a \rightarrow 0$

$y=\sqrt{\alpha +y}; y^{2}=\alpha +y \Rightarrow y^{2}-y-\alpha=0$ $y\quad =\quad \frac { 1\pm \sqrt { 4+\alpha } }{ 2 } \quad \quad (Neglecting\quad -ve\quad sign,\quad as\quad y\quad can't\quad be\quad -ve)$ $y\quad =\quad \frac { 1 + \sqrt { 4+\alpha } }{ 2 } \quad$ $Now,$ $\quad \quad y\quad =\underset { \begin{matrix} x\rightarrow 0 \\ \alpha \rightarrow 0 \end{matrix} }{ lim } \frac { \left[ \frac { 1+\sqrt { 1+4\alpha } }{ 2 } \quad -1 \right] }{ \frac { { x }^{ 2 } }{ 1-cos\quad x } (1-cos\quad x) } =\underset { \alpha \rightarrow 0 }{ lim } \frac { \sqrt { 1+4\alpha } \quad -1 }{ 2.2.\alpha } \quad \quad \left( As\quad \underset { x\rightarrow 0 }{ lim } \frac { { x }^{ 2 } }{ 1-cos\quad x } =2 \right)$

$\quad =\quad \underset { \alpha \rightarrow 0 }{ lim } \frac { (1+4\alpha )\quad -1 }{ 4\alpha \left( \sqrt { 1+4\alpha } +1 \right) }$

$\quad =\quad \underset { \alpha \rightarrow 0 }{ lim } \frac { 1 }{ \sqrt { 1+4\alpha } +1 } \quad =\quad \boxed{\frac { 1 }{ 2 }}$