For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$? For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$?
My thoughts: We know that $(1-\frac{x}{n})^n\leq e^x$, so the numerator is $\leq e^{2x}$.  So, we can play with $\frac{e^{2x}}{n^p}$.  From here, I am not quite sure what to do.  I would really like to be able to find the supremum over $n$, but I can't really minimize the denominator to be able to replace $n$ with something in terms of $x$, because I only have $n^p$ down there.  On the other hand, I feel like I should be splitting up the integral from $0$ to $1$ and then from $1$ to $\infty$ based on the denominator.  Then, fix some $x$, and just use $p$ integral properties to get that $p\in(0,1)$, but I am not quite sure.  Any help is greatly appreciated!
A quick edit: I realize that I made a big mistake above by overlooking the minus sign, so instead the integrand is bounded above by $\frac{1}{n^p}\leq 1$ as $n\rightarrow \infty$, and so we can use DCT and then treat it like a $p$ integral.
A second edit: For Sangchul Lee, I edited the integral to make the upper bound $n$ so he can expand on how he got his approximation.  Thank you!

Assume that $n \ge 2$.
Let
$$I_n := \int_0^n \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x.$$
Using $\ln(1 + u) \le u$ for all $u > -1$, we have, for all $0 \le x < n$,
\begin{align*}
 \ln\left(1 - \frac{x}{n}\right)
 &= \ln \left(1 - \frac{\sqrt n}{n}\right) + \ln\left(1 - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\right)\\[6pt]
 &\le - \frac{\sqrt n}{n} - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\\
 &= - \frac{x}{n - \sqrt n} + \frac{1}{n - \sqrt n}.
\end{align*}
Thus, we have
\begin{align*}
 I_n &\le \mathrm{e}^{n/(n - \sqrt n)}\int_0^n \mathrm{e}^{- nx/(n - \sqrt n) + x}\, \mathrm{d} x\\
 &= \mathrm{e}^{n/(n - \sqrt n)}
 (\sqrt n - 1)(1 - \mathrm{e}^{-n/(\sqrt n - 1)})\\
 &\le \mathrm{e}^{4}\sqrt n.
\end{align*}
On the other hand, we have, for all $x \in [0, \sqrt n)$,
$$\left(1 - \frac{x}{n}\right) \mathrm{e}^{x/n}
\ge \left(1 - \frac{x}{n}\right)\left(1 + \frac{x}{n}\right)
= 1 - \frac{x^2}{n}\cdot \frac{1}{n}
\ge \left(1 - \frac{x^2}{n}\right)^{1/n}$$
where we have used $\mathrm{e}^u \ge 1 + u$ for all $u\ge 0$,
and Bernoulli inequality $(1 + v)^r \le 1 + vr$ for all $0 < r \le 1$ and $v > -1$.
Thus, we have
\begin{align*}
 I_n &\ge \int_0^{\sqrt n} \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x\\
 &\ge \int_0^{\sqrt n} \left(1 - \frac{x^2}{n}\right)  \,\mathrm{d} x\\
 &= \frac23 \sqrt n.
\end{align*}
Thus, we have
$$\frac23 \sqrt n \le I_n \le \mathrm{e}^4 \sqrt{n}, \quad \forall n\ge 2.$$
Thus, $\lim_{n\to \infty} \frac{I_n}{n^p} = 0$ if and only if $p > 1/2$.
