 Solution

Let $e_k=x_k-x_{\ast }$, with $f^{\prime }(x_{\ast })=0$, $f^{\prime \prime }(x_{\ast })\ne 0$. Taylor expansion givesDivision in the Newton update yieldsThus convergence is generically quadratic when that coefficient is nonzero. With a locally Lipschitz second derivative and nonzero $f^{\prime \prime }(x_{\ast })$, the weaker estimate $|e_{k+1}|\le C|e_k{|}^2$ also proves local quadratic convergence.

“Quadratic at best” is false as a universal upper bound. If the displayed coefficient vanishes, the order can be higher. For example $f(x)=x^2/2+x^4/4$ has a minimum at zero, and its Newton optimization update is $x_{k+1}=2x_k^3/(1+3x_k^2)$, which has cubic convergence. For a nondegenerate quadratic objective the method reaches the stationary point in one step. The expansion proves the standard generic quadratic statement while retaining and correcting the source's stronger wording explicitly.