Let \( \Large x = \sqrt{1+\sqrt{1+\sqrt{1+ .....} }} \)
Therefore, \( \Large x = \sqrt{1+x} \)
Squaring both sides, we get
\( \Large x^{2} = 1 + x \)
=> \( \Large x^{2} - x - 1= 0 \)
Therefore, \( \Large x = \frac{1\pm \sqrt{5}}{2} \)
Hence x value lies between 1 and 2 or x vlaue is zero.