 Solution

Define the relative reciprocal error ${\epsilon }_k=1-p{x}_k$. The iteration has the exact identitysoConsequently $|1-p{x}_0|<1$ guarantees $x_k\to 1/p$. For real $p,x_0$, this condition is $0<p{x}_0<2$. A starting guess can be found without division by choosing $x_0$ with the sign of $p$ and repeatedly halving its magnitude until $0<p{x}_0<2$.

For the absolute signed error $e_k=x_k-1/p$, the same identity givesThus the method has cubic convergence, of order of convergence three, with asymptotic error constant $p^2>0$ for a nonexact convergent start. Convergence requires a suitable guess: $p{x}_0=0$ and $p{x}_0=2$ correspond to ${\epsilon }_0=1$ and ${\epsilon }_0=-1$, respectively, so both are fixed failures. For $|1-p{x}_0|>1$ the error magnitude grows without bound. The exact initial value $p{x}_0=1$ is already the answer.