Rudin Real and Complex Analysis Chapter 2 Exercise 15 Generalization How would we compute the limit as $n \to \infty$ of : 
$$\int_{0}^{n} \left(1- \frac{x}{n}\right)^{n} e^{-ax} dx .$$
Rudin asks us to do the cases of $a= 2$ and $a= -\frac{1}{2}$ but I am trying to get more general approach for all $a \in \mathbb R$. 
 A: Assume that $a>-1$ so that the integral actually converges.
It can be seen that for $0<x<n$, we have
$$e^{-x}-\frac1n<\left(1-\frac xn\right)^n<e^{-x}$$
So,
$$\int_0^ne^{-(a+1)x}-\frac{e^{-ax}}ndx<\int_0^n\left(1-\frac xn\right)^ne^{-ax}dx<\int_0^ne^{-(a+1)x}dx$$
Evaluating the left integral, we have
$$\int_0^ne^{-(a+1)x}-\frac{e^{-ax}}ndx=\frac1{a+1}\left(1-e^{-(a+1)n}\right)+\frac1{an}\left(e^{-an}-1\right)\stackrel{n\to\infty}\to\frac1{a+1}$$
On the right integral, we have
$$\int_0^ne^{-(a+1)x}dx=\frac1{a+1}\left(1-e^{-(a+1)n}\right)\stackrel{n\to\infty}\to\frac1{a+1}$$
So by the squeeze theorem,
$$\lim_{n\to\infty}\int_0^n\left(1-\frac xn\right)^ne^{-ax}dx=\frac1{a+1}$$
which was to be expected, since $\lim_{n\to\infty}\left(1-\frac xn\right)^n=e^{-x}$
A: Even though an answer has been accepted, I had a hard time wrapping my head around the inequality $e^{-x}-\frac1n<\left(1-\frac xn\right)^n$. I couldn't prove it however hard I tried. Instead I think above problem can be solved by using Lebesgue's dominated convergence theorem $(LDCT)$.
We have :
$$1-x < e^{-x}, x \in ]0,1[ \\
\implies 1-\frac{x}{n} < e^{-\frac{x}{n}}, x \in ]0,n[\\
\implies \left(1-\frac{x}{n}\right)^n < e^{-x}, x \in ]0,n[ \\
\implies \chi_{[0,n]} . \left(1-\frac{x}{n}\right)^n \leq e^{-x} \
, x \in [0,\infty[ \\
\implies \chi_{[0,n]} . \left(1-\frac{x}{n}\right)^n . e^{-ax} \ \leq e^{-(a+1)x}, x \in [0,\infty[$$
We further have:
$$\int_0^\infty e^{-(a+1)x} dx = \
  \lim_{b \to \infty}\left(\frac{1-e^{-(a+1)b}}{a+1}\right) \
  = \frac{1}{a+1} < \infty$$
Therefore $e^{-(a+1)x} \in L^1([0,\infty])$. Applying $(LDCT)$ we get:
$$\lim_{n \to \infty} \int_0^n \
  \left(1-\frac{x}{n}\right)^n . e^{-ax} \\
  = \lim_{n \to \infty} \int_0^\infty \chi_{[0,n]} . \ 
  \left(1-\frac{x}{n}\right)^n . e^{-ax} dx \\
  = \int_0^\infty e^{-(a+1)x} dx = \frac{1}{a+1}$$
