Find the limit of convergence sequence

Let $x_1=1$ and $x_{n+1}=\sqrt{1+2/x_n}$, for $n\in\mathbb{N}$. Show that $\lim\limits_{n\rightarrow\infty} x_n$ exists and find the limit.

I have conjectured that odd-term sequence of $x_n$ is increasing and even-term sequence of $x_n$ is decreasing, but I am struggling to prove that $x_{2n+1}-x_{2n-1}>0$ and $x_{2n}-x_{2n-2}<0$, for all $n$.

I have conjectured that odd-term sequence of $x_n$ is increasing and even-term sequence of $x_n$ is decreasing, but I am struggling to prove that $x_{2n+1}-x_{2n-1}>0$ and $x_{2n}-x_{2n-2}<0$

Hint: $\require{cancel}\;x_{n+1}^2-x_n^2 = \left(\cancel{1}+\dfrac{2}{x_n}\right)-\left(\cancel{1}+\dfrac{2}{x_{n-1}}\right)=\dfrac{-2(x_n-x_{n-1})}{x_nx_{n-1}}\,$, therefore the difference between consecutive terms changes sign at each step. It follows that the subsequences of odd, and respectively even, indices are each monotonic, and of opposed monotonicity.

You can also see it as a fixpoint problem for $f(x) = \sqrt{1+\frac{2}{x}}$ on $[1,2]$ because $$f:[1,2]\rightarrow [1,2] \mbox{ and } |f´(x)| = \left|-\frac{1}{x^2\sqrt{1+\frac{2}{x}}} \right| <= \frac{1}{\sqrt{3}}<1 \mbox{ on } [1,2]$$ So, the limit exists and is the only real solution to $x^3-x-2=0$.