# Help with Real Analysis Integral

I'm working through practice problems and I came along the following:

Evaluate $\lim_{n\rightarrow\infty}\int_0^n(1-\frac{x}{n})^n dx.$

I think this should work out via Dominated Convergence Theorem, but I can't seem to get it. I have been starting with $x\in(0,n)\Rightarrow 0<(1-\frac{x}{n})<1 \Rightarrow (1-\frac{x}{n})^n\leq(1-\frac{x}{n}).$ I just don't see how to use this. I can't get a bound for the integral that doesn't rely upon $n.$

Am I coming at it from the wrong angle? Any assistance is appreciated.

For sufficiently large $n$, we have $$\int_0^n\left(1-\frac{x}{n}\right)^ndx=\frac{(x-n)(1-\frac{x}{n})^n}{n+1}\Bigg|_0^n=\frac{n}{n+1}.$$ What is the limit of that as $n$ goes to infinity?
Show that the following inequality is always satisfied: $$\left(1-\dfrac{x}n\right)^n \mathbb{1}_{x \in [0,n]} \leq e^{-x}$$ Then use Lebesgue dominated convergence theorem.
You can't use dominated convergence because of the $n$ in the upper limit. Why not just compute the integral directly?
$$\int_0^n \left(1-\frac{x}{n}\right)^n dx=\frac{n}{n+1}$$
So $$\lim\limits_{n\to\infty} \int_0^n \left(1-\frac{x}{n}\right)^n dx=1$$