Limits in Banach Spaces Let $X$ be a Banach space and $I$ be the identity operator in $X$. I am trying to prove that $$\lim_{t\to 0^+} \frac{e^t I - I}{t} = I,$$ in $L(x)$. This is the same as proving that, for any $x\in X$, $$\lim_{t\to 0^+} \dfrac{e^t x - x} t = x.$$ What I did at first was $$\lim_{t\to 0^+} \frac{e^t x - x}t  = x \lim_{t\to 0^+} \frac{e^t - 1} t = x,$$ but this does not seem right to me... any hints? Thank you very much!
 A: Any statement about the limit of an operator-valued function depends on which topology on the set of operators one has in mind. One can define for operators $A$ the norm $$\|A\| = \sup_{x\,\in\, X\, \smallsetminus \,\{0\}} \frac{\|Ax\|}{\|x\|} $$ where $x\mapsto \|x\|$ is the Banach space's norm, and then $A\to B$ means $\|A-B\| \to 0.$ The question then is whether
$$
\lim_{t\,\downarrow\,0} \left( \frac{e^t-1} t I \right) = \left( \lim_{t\,\downarrow\,0} \frac {e^t - 1} t \right) I \text{ ?}
$$
Since the limit on the right is a limit of a real-valued function and is equal to $1,$ we ask whether
$$
\lim_{t\,\downarrow\,0} \left\| \frac{e^t-1} t I - I \right\| =0,
$$
where this is now a limit of a real-valued function.
From basic identities about normed spaces we get this:
$$
\frac{\left\| \left( \frac{e^t - 1} t I - I \right) x \right\|}  {\left\| x \right\|} = \frac{\left\| \left( \frac{e^t - 1} t - 1 \right) I x \right\|}  {\left\| x \right\|} = \frac{\left| \frac{e^t-1} t - 1 \right|\|x\|}{\|x\|} = \left| \frac{e^t - 1} t - 1 \right|.
$$
Thus sup over all $x\in X$ of the expression on the left cannot exceed the expression on the right.
