Truncated Taylor series of the exponential Let $N \in \mathbb N^*$, $\delta > 0$, $t > 0$ and consider
\begin{equation}
f(t,\delta, N)= e^{-t/\delta} \sum_{k=0}^{N-1}\frac{(t/\delta)^k}{k!}. \qquad \qquad \qquad (1)
\end{equation}
Let now $N = \lceil\delta^{-\gamma}\rceil$, with $0 < \gamma < 1$. Can we conclude that
\begin{equation}
\lim_{\delta \to 0} f(t,\delta,\lceil \delta^{-\gamma}\rceil)=0,
\end{equation}
for all $t > 0$?
The reasoning is that if $N \to \infty$ (for $\delta$ fixed), then the sum tends to $e^{t/\delta}$ and therefore $f(t,\delta,N)\to 1$. If on the other hand $\delta \to 0$, for $N$ fixed, we have that $f(t,\delta,N)\to 0$ since the exponential goes faster to zero than the polynomial to infinity. In this case, we let $\delta \to 0$ and simultaneously $N \to \infty$, but with a "slower pace". Can we conclude that the limit is zero? If yes, how?
Edit:
After some reasoning I came up with
$$
f(t,\delta,N)= \frac{\Gamma(N, t/\delta)}{(N-1)!},
$$
with $\Gamma(\cdot, \cdot)$ the upper incomplete Gamma function. I implemented this in Matlab, and the interesting behaviour is that if $N=\delta^{-1}$, then the limit (1) is $1/2$ if $0 <t \leq 1$ and it is 0 if $t > 1$. If $N=\delta^{-\gamma}$, with $\gamma \in (0, 1)$, the limit is 0, and if $N = \delta^{-\gamma}$, with $\gamma > 1$, then the limit is 1, independently of $t$.
 A: For simplicity let $x=t/\delta$ for the moment.
The main point is that $\frac{x^k}{k!}$ is within a polynomial factor of $\left ( \frac{xe}{k} \right )^k$, by Stirling. (It's within a bounded factor of $\frac{(xe)^k}{k^{k+1/2}}$ but this is overkill for this application.)
So for $0<c<1$, the sum of all the terms with $k>x(1+c)$ behaves basically like $\left ( \frac{e}{1+c} \right )^{x(1+c)} \frac{c+1}{c}$. This is the dominant term times $\frac{1}{1-r}$ where $r=\frac{1}{1+c}$ is the common ratio of the bounding geometric series.
Similarly for the first $x(1-c)$ terms you have a bound like $\left ( \frac{e}{1-c} \right )^{x(1-c)} \frac{1}{c}$.
Returning to your setting, if $\gamma>1$ then you exclude only the terms past $x(1+c)$ where $c \to \infty$ as $\delta \to 0$, and so the error goes to zero.
If $\gamma<1$ then you exclude all the terms past $x(1-c)$ for $c \to 1$ as $\delta \to 0$ and so the terms you are including become negligible relative to $e^x$, i.e. the error goes to $1$.
If instead you look at $N \sim C \delta^{-1}$ then the above bounds properly come into play.
