Question on the use of a limit in a proof

I ran into a step in an argument that I can't quite figure out. It's basically how they use a limit that I don't seem to understand. The context is local-to-unity asymptotics in vector autoregressions, so I figured some people doing empirical work might know.

We have a $$K \times K$$ diagonal matrix $$\Lambda := \text{diag}\left( \lambda_1, \dots, \lambda_K \right)$$ and we define $$h := \left[ \delta T \right]$$ where $$[.]$$ refers to the integer part of a number and $$\delta \in (0,1)$$ is some fraction. We'd like to know what happens to $$\Lambda^h$$ as $$T \rightarrow \infty$$. The textbook notes that \begin{align*} \lim_{T \rightarrow \infty} \left( 1 + \frac{\delta \lambda_k}{T} \right)^T = e^{\delta \lambda_k} \; \forall k=1,\dots,K \end{align*} and infers that $$\Lambda^h \rightarrow e^{\delta \Lambda} := \text{diag}\left( e^{\delta \lambda_1}, \dots, e^{\delta \lambda_K} \right)$$. I don't quite follow what is going on. I see that the first limit is correct, but I fail to see how the conclusion follows. If it helps, later the authors actually use the reasoning to claim that $$C^h \rightarrow e^{\delta C}$$, but $$C := I_K + \Lambda/T$$. Now, in that case we have $$\left( C^h \right)_{k,k} = \left( 1 + \frac{\lambda_k}{T} \right)^{[\delta T]}$$. I still am having trouble seeing how the hell $$\delta$$ can be brought inside to invoke the above argument, but the expression is closer.

Anyone can help me out here?

It makes no sense to write $$C^h \to e^{\delta C}$$ as $$T \to \infty$$ when $$C = I_K +\Lambda/T$$ since $$e^{\delta C}$$ on the right-hand side depends on $$T$$.

What can be confirmed is $$(C^h)_{k,k} \to e^{\delta \lambda_k}$$ as $$T \to \infty$$ with $$\delta$$ fixed. Note that

$$\log (C^h)_{k,k} = \lfloor \delta T\rfloor \log\left(1 + \frac{\lambda_k}{T}\right) = \frac{\lfloor \delta T\rfloor}{T} \log\left(1 + \frac{\lambda_k}{T}\right)^T$$

As $$T \to \infty$$ we clearly have $$\log\left(1 + \frac{\lambda_k}{T}\right)^T \to \log e^{\lambda_k} = \lambda _k$$, since $$\log(\cdot)$$ is continuous.

Since $$\lfloor \delta T\rfloor \leqslant \delta T < \lfloor \delta T\rfloor +1$$, it follows that $$\delta T -1 < \lfloor \delta T\rfloor \leqslant \delta T$$, and $$\delta - \frac{1}{T} < \frac{\lfloor \delta T\rfloor}{T} \leqslant \delta$$.

Hence, by the squeeze theorem, $$\frac{\lfloor \delta T\rfloor}{T} \to \delta$$ as $$T \to \infty$$, and

$$\lim_{T \to \infty}\log (C^h)_{k,k} = \lim_{T \to \infty}\frac{\lfloor \delta T\rfloor}{T} \lim_{T \to \infty}\log\left(1 + \frac{\lambda_k}{T}\right)^T = \delta \lambda_k$$

Therefore, by continuity of $$\log(\cdot)$$,

$$\lim_{T \to \infty} (C^h)_{k,k} = e^{\delta \lambda_k}$$