Which constant is it?

If we observe the problem closely,we can see that the whole question is embedded in itself.So we can write the problem as: $x=1+\frac{1}{x}$ Rearranging this, we get: $x=1+\frac{1}{x}\rightarrow x^2-x-1=0$ Putting in the respective values in the quadratic formula,we get; $\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}$ $\frac{1\pm\sqrt{1+4}}{2}$ $\frac{1\pm\sqrt{5}}{2}$ Now as the infinite continued fraction is positive,we discard the negative root and the answer is $\frac{1+\sqrt{5}}{2}$ .Now this number is a very special number;it is... $\LARGE{The\;Golden\;Ratio!!!}$ Which is denoted by $\phi$ so answer is $\boxed{\phi}$

The fraction doesn't ever change, so that means we can say $x=1+\frac{1}{x}$

Multiplying by $x$ on both sides gives us $x ^ 2=1+x$ .

Subtracting $x$ and then $1$ from both sides will then give us the quadratic $x ^ 2-x-1=0$ .

From this, we know $a=1$ , $b=-1$ , and $c=-1$ .

Plugging this into the quadratic formula for x becomes this: $x=\frac{1\pm\sqrt{5}}{2}$ . If we choose to add 1, we get this:

$x=\frac{1+\sqrt{5}}{2}$

Sound familiar? It should, because the above is also equal to phi, so $\phi$ is the answer!