Prove that $X \in \mathfrak g.$ Let $E$ be a Banach space. Let $G \subseteq GL(E)$ be a closed subgroup i.e. $G$ is a linear Banach Lie group. Define $$\mathfrak g : = \left \{X \in \mathcal L(E)\ \big |\ \forall t \in \Bbb R, e^{tX} \in G \right \}.$$ Let $\gamma : (-\varepsilon, \varepsilon) \longrightarrow G$ be a $C^1$-map with $\gamma (0) = I.$ Show that $X = \dot {\gamma} (0) \in \mathfrak g.$

By Taylor's theorem I get for all $t \in (-\varepsilon, \varepsilon)$ $$\gamma (t) = I  + t X + o (t).$$
From here how do I conclude that $e^{tX} \in G,$ for all $t \in \Bbb R\ $? Would anybody please help me in this regard?
Thanks in advance.
EDIT $:$ If $G = GL(E)$ then we are through since $e^T \in GL(E)$ if $T \in \mathcal L (E).$
Lemma $:$ Let $\{A_n \}_{n \geq 1}$ be a sequence in $\mathcal L(E)$ with $\lim\limits_{n \to \infty} A_n = A.$ Then $$\lim\limits_{n \to \infty} \left (I + \frac {A_n} {n} \right )^n = e^A.$$
Let us prove the required result with the help of the above lemma.
Let $t \in \Bbb R.$ Then there exists $N \in \Bbb N$ such that for all $n \geq N$ we have $\left \lvert \frac {t} {n} \right \rvert \lt \varepsilon$ i.e. $\frac {t} {n} \in (-\varepsilon,\varepsilon).$ Hence it follows that $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N.$ Therefore for all $n \geq N$ we have $\gamma \left (\frac {t} {n} \right )^n \in G,$ since $G$ is a group. Now by Taylor's theorem we have $$\begin{align*} \gamma \left (\frac {t} {n} \right ) &amp; = \gamma (0) + \frac {t} {n} \dot {\gamma} (0) + o \left (\frac {t} {n} \right ) \\ &amp; = I + \frac {t} {n} X + o \left (\frac {t} {n} \right )\ \ (\because \gamma (0) = I\ \text {and}\ \dot{\gamma} (0) = X\ ) \end{align*}$$ Hence from the above lemma it follows that $$\lim\limits_{n \to \infty} \gamma \left (\frac {t} {n} \right )^n = \lim\limits_{n \to \infty} \left (I + \frac {tX + o(1)} {n} \right )^n = e^{tX}\ \ (\because o(1) \to 0\ \text {as}\ n \to \infty\ )$$
Since $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N$ and $G$ is closed it follows that $e^{tX} \in G.$ Since $t \in \Bbb R$ was arbitrarily taken it follows that $e^{tX} \in G,$ for all $t \in \Bbb R.$ Hence $X \in \mathfrak g,$ as required.
QED
Lemma $:$ Let $\{A_n \}_{n \geq 1}$ be a sequence in $\mathcal L(E)$ with $\lim\limits_{n \to \infty} A_n = A.$ Then $$\lim\limits_{n \to \infty} \left (I + \frac {A_n} {n} \right )^n = e^A.$$
Let us prove the required result with the help of the above lemma.
Let $t \in \Bbb R.$ Then there exists $N \in \Bbb N$ such that for all $n \geq N$ we have $\left \lvert \frac {t} {n} \right \rvert \lt \varepsilon$ i.e. $\frac {t} {n} \in (-\varepsilon,\varepsilon).$ Hence it follows that $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N.$ Therefore for all $n \geq N$ we have $\gamma \left (\frac {t} {n} \right )^n \in G,$ since $G$ is a group. Now by Taylor's theorem we have $$\begin{align*} \gamma \left (\frac {t} {n} \right ) &amp; = \gamma (0) + \frac {t} {n} \dot {\gamma} (0) + o \left (\frac {t} {n} \right ) \\ &amp; = I + \frac {t} {n} X + o \left (\frac {t} {n} \right )\ \ (\because \gamma (0) = I\ \text {and}\ \dot{\gamma} (0) = X\ ) \end{align*}$$ Hence from the above lemma it follows that $$\lim\limits_{n \to \infty} \gamma \left (\frac {t} {n} \right )^n = \lim\limits_{n \to \infty} \left (I + \frac {tX + o(1)} {n} \right )^n = e^{tX}\ \ (\because o(1) \to 0\ \text {as}\ n \to \infty\ )$$
Since $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N$ and $G$ is closed it follows that $e^{tX} \in G.$ Since $t \in \Bbb R$ was arbitrarily taken it follows that $e^{tX} \in G,$ for all $t \in \Bbb R.$ Hence $X \in \mathfrak g,$ as required.
QED