5. Let $x_{1}=2, x_{n+1}=\frac{2 x_{n}-1}{x_{n}},(n \geqslant 1)$, then $\lim _{n \rightarrow \infty} x_{n}=$
5. Let $\lim _{n \rightarrow \infty} x_{n}=t$,

then from $x_{n+1}=\frac{2 x_{n}-1}{x_{n}}$,
we get $t=\frac{2 t-1}{t}$,
solving for $t$ yields $t=1$.