Evaluate : $\lim \limits_{n\to\infty}n(1-\int_0^{n}(1-\frac{x}{n})^ndx)$ Evaluate : 
$a)$ $I=\lim \limits_{n\to\infty}n(1-\int_0^{n}(1-\frac{x}{n})^ndx)$
$b)$ $\lim \limits_{n\to\infty}(2-\int_0^{n}(1-\frac{x}{n})^ndx)^n$
About The first limit : 
My idea is : 
$I=\lim \limits_{n\to\infty}\frac{1-\int_0^{n}(1-\frac{x}{n})^ndx}{\frac{1}{n}}$ 
In form
$\frac{0}{0}$ 
I think use stolze Cesaro theorem 
But the second limit I don't have any ideas to approach it!
 A: For part 1, we can just directly evaluate the integral. Don't be tempted by the familiar expression to write $(1-\frac{x}{n})^n = e^{-x}$. It's quite straightforward, using the substitution $u = 1 - \frac{x}{n} \implies du = -\frac{dx}{n}$. So the integral becomes
$$I =\int_0^n \big(1- \frac{x}{n}\big)^n dx = -n \int_1^0 u^n du = \frac{n}{n+1}$$
Then we have $\lim_{n \to \infty} n (1 - \int_0^n \big(1- \frac{x}{n}\big)^n dx) = \lim_{n \to \infty} \frac{n}{n+1} = \boxed{1}$.
For part 2, we use what we calculated in part one and we have $2 - I = \frac{n+2}{n+1}$. Then the limit becomes $L = \displaystyle \lim_{n \to \infty} \bigg( \frac{n+2}{n+1} \bigg)^n \implies \ln L = \lim_{n \to \infty} \frac{(\ln(n+2) - \ln(n+1))}{\frac{1}{n}}$. By L'Hopitale, we have $\ln L = \displaystyle \lim_{n \to \infty} \frac{\frac{1}{n+2} - \frac{1}{n+1}}{-1/n^2} = \lim_{n \to \infty} \frac{\frac{-1}{(n+1)(n+2)}}{-1/n^2} = 1$. So $L = e^{1} = \boxed{e}$.
A: Before we bring the big guns, the change of variables $u=1-\frac{x}{n}$ gives
$$
\int^n_0(1-\frac{x}{n})^n\,dx=n\int^1_0u^ndu=\frac{n}{n+1}
$$
That will simplify things; for instance,
$$n\Big(1-\int^n_0(1-\frac{x}{n})^ndx\Big)=n\Big(1-\frac{n}{n+1}\Big)$$
