Fraction in fraction in fraction ............

This can be written as $\dfrac{1}{1 + x} = x \Longrightarrow x^{2} + x - 1 = 0 \Longrightarrow x = \dfrac{-1 \pm \sqrt{5}}{2}$ .

Clearly $x \gt 0$ , so $x = \dfrac{\sqrt{5} - 1}{2} = \boxed{0.62}$ to $2$ decimal places.

for(int i=0;i<10;i++) { x= 1/(1+x); }

cout<<x;

i did it vai programming

The given expression can be written as : $\frac{1}{1+x}$ because , if you remove one element from an infinite set of things, that won't change the infinity much. So here we assume the things after the number 1 in the denominator to be x again(by virtue of above property ).

Rest solution: $\frac{1}{1+x} =x$ 1=(\x^{2}) +x So, $\x=\frac{-1+5^{0.5}}{2}$ since\ (\sqrt{5}=2.23) $\x=\frac{-1+2.23}{2}$ $\x=0.618$

https://www.xtremepapers.com/community/attachments/sat-work-png.49171/

x = 1/(1+x) x = (sqrt 5 - 1)/2 = 0.618, x = - (sqrt 5 + 1)/2 = -1.618 x = 0.62 or x = -1.62

$simply\quad \frac { \sqrt { 5 } -1 }{ 2 }$