So, the setup is as follows:
Let $X$ be a Banach space, and let $L \in \mathcal{B}(X,X)$ be a bounded linear operator. For every $t \in \mathbb{R}$ and $k \geq 0$, define $$E_k(t) := \sum_{j=0}^k \frac{t^j}{j!}L^j$$ where $L^j$ is just the $j$-fold composition of $L$ with itself. I am given the following tasks:
1. Show that for every $t \in \mathbb{R}$, $$\lim_{h \to 0} \sup_{k \in \mathbb{N}} \frac{||E_k(t+h)-E_k(t) - hLE_{k-1}(t)||}{|h|} = 0$$ so that, in particular, $E_k(t)$ is differentiable in $t$ with $$E'_k(t) = \lim_{h \to 0} \frac{E_k(t+h) - E_k(t)}{h} = LE_{k-1}(t)$$ 2. Show that $\{E_k(t)\}_k$ is a Cauchy sequence in $\mathcal{B}(X,X)$. Then, since $\mathcal{B}(X,X)$ is Banach, denote its limit with $E(t)$ and show, using part 1, that $E'(t)$ exists and is equal to $LE(t)$.
So far, I've been able to tackle 2. a little bit and have proven the sequence given is Cauchy. Then, to show the existence of $E'(t)$, I'd like to be able to do the following:
$$\lim_{h \to 0} \frac{E(t+h) - E(t)}{h} = \lim_{h \to 0} \left ( \lim_{k \to \infty} \frac{E_k(t+h) - E_k(t)}{h} \right) \underbrace{=}_{\text{*}} \lim_{k \to \infty} \left( \lim_{h \to 0} \frac{E_k(t+h) - E_k(t)}{h} \right) = \lim_{k \to \infty} LE_{k-1}(t) \underbrace{=}_{\text{since $L$ is bounded, hence continuous}} L \left(\lim_{k \to \infty} E_{k-1}(t) \right) = LE(t)$$
My questions are, how can I justify the equality $*$ (if this is even the right path to go down), and I'm a bit lost on $1.$ entirely, so any help would be appreciated!