Find (with a proof ) lim$_{n \rightarrow \infty} \int_0^n \frac{(1 -\frac{x}{n})^n}{ne^{-x}}dx$ So I am pretty sure I want to start off with defining $f_n = \frac{(1 -\frac{x}{n})^n}{ne^{-x}}$ if $x >0$, and then use something like the bounded convergence theorem or dominated convergence theorem.

 A: Use the substitution $u = 1 - \frac{x}{n}$ to get
$$\lim_{n\to\infty}\int_0^1\left(ue^{1-u}\right)^n\:du$$
Now the key to this problem is taking a look at the integrand on the interval $(0,1)$
$$\frac{d}{du}(ue^{1-u}) = (1-u)e^{1-u} = 0 \implies u = 1$$
And since the integrand vanishes at $u=0$, this means $u=1$ is a max and the function is less than $1$ on $(0,1)$, meaning
$$\lim_{n\to\infty}\int_0^1\left(ue^{1-u}\right)^n\:du \to \int_0^1 0\:du = 0$$
by dominated convergence. Now had the original integral not had the $n$ in the denominator, we would have gotten something more interesting, namely
$$\lim_{n\to\infty}\int_0^n\frac{\left(1-\frac{x}{n}\right)^n}{e^{-x}}\:dx = 1$$
A: By a simple change of variables, we have
$$
I_n := \int_0^n {\frac{{\left( {1 - \frac{x}{n}} \right)^n }}{{ne^{ - x} }}dx}  = \int_0^1 {\left[ {(1 - t)e^t } \right]^n dt} .
$$
Now,
$$
0 \le (1 - t)e^t  = 1 - \frac{{t^2 }}{2}e^\xi  (1 + \xi ) \le 1 - \frac{{t^2 }}{2} \le e^{ - t^2 /2} 
$$
with a suitable $\xi \in (0,1)$ (Taylor's formula with Lagrange remainder). Hence,
$$
0 \le I_n  \le \int_0^1 {e^{ - nt^2 /2} dt} .
$$
By the dominated convergence theorem, the right-hand side tends to $0$ and thus $I_n \to 0$. If you do not want to use dominated convergence, note that
$$
I_n  \le \int_0^1 {e^{ - nt^2 /2} dt}  \le \int_0^{ + \infty } {e^{ - nt^2 /2} dt}  = \sqrt {\frac{\pi }{{2n}}} .
$$
A: You could obtain much more than the limit. Considering $$I_n=\int_0^n\frac{ \left(1-\frac{x}{n}\right)^n}{n}\,e^x\,dx$$ let $x=nt$ to make
$$I_n=\int_0^1 (1-t)^n \,e^{n t}\,dt$$
$$J_n=\int (1-t)^n \,e^{n t}\,dt=e^n \, n^{-(n+1)}\,\Gamma (n+1,n(1-t))$$
$$I_n=e^n \, n^{-(n+1)}\,\Big[\Gamma (n+1) -\Gamma (n+1,n) \Big]$$
$$I_n=e^n \, n^{-(n+1)}\,\Gamma (n+1)\Big[1 -\frac{\Gamma (n+1,n)}{\Gamma (n+1)}  \Big]$$ When $n$ is very large
$$\Big[1 -\frac{\Gamma (n+1,n)}{\Gamma (n+1)}  \Big]\to \frac 12$$ For the remaining, use logarithms and Stirling approximation, to get
$$\log\Big[e^n \, n^{-(n+1)}\,\Gamma (n+1)\Big]=\frac{1}{2} \log \left(\frac{2 \pi }{n}\right)+\frac{1}{12 n}-\frac{1}{360n^3}+O\left(\frac{1}{n^5}\right)$$
$$I_n= \sqrt{\frac{ \pi}{2n}} \exp \left(\frac{1}{12 n}-\frac{1}{360n^3}+\cdots \right)$$ which shows the limit and how it is approached.
Trying for $n=10^6$, the exact value is $0.00125265$ while the approximation gives $0.00125331$.
