I'm studying constrained optimization. For instance, consider the EQ constrained optimization problem of the form
\begin{align} min_{x \in X} && f(x)\\ s.t && h_i(x)=0 && i=1,2,...m &&&&&& (1) \end{align}
So what I so far understood is that the idea of the penalty method is to eliminate the constraints and add a penalty term to the objective function and transform the constrained minimization (1) to the following unconstrained minimization
\begin{align} min_{x \in \mathbb{R}^n} && P(x;r)=f(x) + \frac{r}{2}h(x)^T h(x) && r>0 && (2) \end{align}
so now (2) is an approximation of (1) so in order for (1) and (2) to be equivalent when the first-order necessary conditions coincide
First-order necessary conditions for (1)
let $l(x, \lambda)=f(x) + \lambda^T h(x)$ $\implies$
$\nabla f+ \nabla h=0$
$h(x)=0$
First-order necessary condition for (2)
$\nabla f + r \nabla h h=0$
and they will coincide when $rh$ behaves like $\lambda$ and $r \to \infty$ which implies that $h \to 0$
QUESTION I want to know the meaning of the last 2 lines. As far as I understood is that we want $r$ to get very big so that (2) approaches (approximates well) (1) but also we want $rh$ to behave like $\lambda$ but because $\lambda$ is finite and we already want $r \to \infty$ then $h \to 0$
Is that correct?