First hint:
It’s $1/φ$. Check out the golden ratio.
Second hint, which adds a bit to answers given so far:
Since you should probably apply limits formally, you can describe it as the limit of a sequence given by $a_0 = 1$ and $a_{k+1} = \frac{1}{1 + a_k}$. Now if $x = \lim_{k → ∞} a_k$, then
$$\frac{1}{1+x} \overset{\text{limit rules}}{=} \lim_{k → ∞} \frac{1}{1 + a_{k}} = \lim_{k → ∞} a_{k+1} = x$$
So $x^2 + x - 1 = 0$.
To show that limit exits in the first place, you can tell from the frist few members:
$$ 1,\; \frac{1}{2},\; \frac{2}{3},\; \frac{3}{5},\; \frac{5}{8},\; \frac{8}{13},\; …$$
that the sequence is actually given by
$$ a_k = \frac{f_k}{f_{k+1}};\quad \text{where}\; f_{k+2} = f_{k+1} + f_k, \text{and}\; f_1 =f_2 = 1,$$
meaning $f_k$ denotes the $k$-th Fibonacci number, and readily prove this by induction.
Now, the $k$-th Fibonacci number is given by $f_k = \frac{φ^k - ψ^k}{\sqrt{5}}$ where $φ = \frac{1+\sqrt{5}}{2}$ and $ψ = \frac{1 - \sqrt{5}}{2}$, the roots of $X^2 - X - 1$. This is true since the Fibonacci sequence is uniquely determined by the recursion and its first two members.
But that foregoing condition implies $φ^{k+2} = φ^{k+1} + φ^k$ as well as $ψ^{k+2} = ψ^{k+1} + ψ^k$, so:
$$\frac{φ^{k+2} - ψ^{k+2}}{\sqrt{5}} = \frac{φ^{k+1} - ψ^{k+1}}{\sqrt{5}} + \frac{φ^k - ψ^k}{\sqrt{5}},$$
whereas $\frac{φ^1 - ψ^1}{\sqrt{5}} = \frac{φ^2 - ψ^2}{\sqrt{5}} = 1$.
Now $\frac{f_k}{φ^k} \overset{k → ∞}{\longrightarrow} \frac{1}{\sqrt{5}}$, since $0 < ψ < 1$ and so $ψ^k \overset{k → ∞}{\longrightarrow} 0$ and $0 < \frac{1}{φ} < 1$ and so $\frac{1}{φ^k} \overset{k → ∞}{\longrightarrow} 0$.
This, however, implies:
$$\frac{f_k}{f_{k+1}} = \frac{1}{φ} · \frac{f_k}{φ^k}·\frac{φ^{k+1}}{f_{k+1}} \overset{k → ∞}{\longrightarrow} \frac{1}{φ}.$$
