An arithmetico-geometric sequence

1. $\ell=\frac12\ell+1\iff\ell=2$.

2. Subtracting $\ell=2$ from $u_{n+1}=\frac12u_n+1$ gives
$$v_{n+1}=u_{n+1}-2=\frac12(u_n-2)=\frac12v_n.$$
So $(v_n)$ is geometric with ratio $\frac12$.

3. $v_0=-2$, so $v_n=-2\cdot2^{-n}=-2^{1-n}$ and
$$u_n=2-2^{1-n}\longrightarrow2.$$

Remark. The trick of subtracting the fixed point works for every recurrence $u_{n+1}=au_n+b$ with $a\ne1$: it turns it into a geometric sequence of ratio $a$.