Let $(T_t)_{t\geq 0}$ be a $C_0$-semigroup on a Banach space $X$ with generator $A$ such that the spectral bound $s(A)=0.$ Suppose there exists an operator $P$ on $X$ such that $$T(t) \stackrel{t\to \infty}{\to} P \text{ strongly }.$$
Then I was able to show that $$\lim_{\lambda\to 0}\lambda R(\lambda,A)f \text{ exists for each } f\in X \qquad\qquad (1).$$ I was told the converse is true if the semigroup is holomorphic.
However, I'm wondering what can be said about the limit in $(1).$
I know that if $0$ is an isolated spectral value and a pole, then the limit in $(1)$ is the spectral projection associated to $0.$
What happens when it is not an isolated spectral value? Is the limit in $(1)$ equal to $P?$
Edit: Let $\lambda>0.$ Since $T(t) \to P,$ therefore $\mathrm{Im}\, P=\ker A$ and $\overline{\mathrm{Im}\, A}\subseteq \ker P.$ The first implies that $$\lambda R(\lambda,A)P=P$$ and the second implies that $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=0 \text{ on a closed subspace of } \ker P.$$
Can we now conclude by the existence of limit that, infact $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=0 \text{ on } \ker P$$ and hence $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=P?$$
Part 1 of the answer.
Yes, if the semigroup converges strongly to $P$, then $\lambda R(\lambda,A)$ converges strongly to $P$, too, as $\lambda \downarrow 0$. (You do not need the assumption $s(A) = 0$ for this.)
This follows from the Laplace transform representation of the resolvent, i.e., from the formula $$ \lambda R(\lambda,A)f = \int_0^\infty \lambda e^{-t\lambda}T_t f \, dt \quad \text{for all } f \in X, $$ which is true for each complex $\lambda$ with real part $> 0$ (since convergence of the semigroups implies, by means of the uniform boundedness theorem, that its growth bound $\le 0$).
Part 2 of the answer.
This is not correct. The one-dimensional semigroup $(e^{it})_{t \ge 0}$ is a counterexample.