I am referring to this paper, p. 26., Theorem 3.8.
In the proof of this Theorem, it is said that in order to deduce from the statement $$ \forall n\in\mathbb{N}\qquad\mathcal{H}(f^n)\leq (1+2\kappa\delta t)^{-n}\mathcal{H}(f^0) $$ that, for all $\delta t>0$, there exists $k>0$ (explicit) with $\lim_{\delta t\to 0}k=\kappa$, such that $$ \forall n\in\mathbb{N}\qquad\mathcal{H}(f^n)\leq e^{-2kn\delta t}\mathcal{H}(f^0) $$
one just needs to do a Taylor development of the exponential function. How is that meant?
As far as I see, the role of $\mathcal{H}$ does not play any role for my question (this is why I do not mention its very lengthy definition here).
I do not have an exact idea. But I am pretty sure one needs $$ 1+2\kappa\delta t + O((\delta t)^2)=e^{2\kappa\delta t},\quad\textrm{ as }\delta t\to 0, $$ where $O(\cdot)$ is the Big-O notation. Thus, if one defines some $k=k(\delta t)=\kappa + \varphi$ with $\varphi=\varphi(\delta t)\to 0$ as $\delta t\to 0$, one should have, as $\delta t\to 0$, $$ 1+2\kappa\delta t = e^{2k\delta t}\tag{1} $$
I am note sure if this is thought too naively. Particularly, I am not sure if for each fixed $\delta t>0$, equation (1) holds; I think if $\delta t>0$ is small enough, (1) holds if one chooses $\varphi$ appropriately.
I'm assuming all things are positive.
I'm sure that a Taylor argument like yours should work out fine, but you can also just define $$ k = \frac{\log(1+2\kappa\delta t)}{2\delta t}. $$ Then $k>0$, $ e^{2k\delta t} = 1+2\kappa\delta t, $ and $ \lim_{\delta t\to0^+} k = \kappa $.
I think to finish your argument, you would just need to define your $\varphi$ such that $e^{2k\delta t} = 1+2\kappa\delta t$ holds, and then show the desired properties from there.
I'll add a way to finish your Taylor argument. There are probably slightly sleeker ways than what follows, but here goes.
For simplicity of notation, let $h=2\delta t$. Choose a function $\varphi(h)$ such that $$ \tag{$\ast$} e^{(\kappa+\varphi(h))h} = 1+\kappa h. $$ Can we do this? Yes, just solve the equation to find a valid $\varphi$ (it's similar to how I defined $k$ earlier). Define $k(h) = \kappa + \varphi(h)$.
From $(\ast)$, it follows that $h\to0$ implies $(\kappa+\varphi(h))h\to0$. Then, $$ 1+\kappa h = e^{(\kappa+\varphi(h))h} = 1 + (\kappa+\varphi(h))h + o_{h\to0}((\kappa+\varphi(h))h), $$ where I use little-oh notation. From that equation, we get $\varphi(h)h = o((\kappa+\varphi(h))h),$ which implies (use the definition of little oh) that $\varphi(h)=o(1)$, which is what we wanted. Oh, and also $k>0$ for $h>0$ follows from $(\ast)$.