How to verify this limit? Kindly asking for any hints about showing:
$$\lim_{n\to\infty}\int_0^1\frac{dx}{(1+x/n)^n}=1-\exp(-1)$$
Thank you very much, indeed!
 A: Well by substitution, $u=1+ \frac{x}{n}$, $du=\frac{dx}{n}$,
$$\int\frac{dx}{(1+\frac{x}{n})^n}=n\int\frac{du}{u^n}=\frac{n}{1-n}u^{-n+1}=\frac{n}{1-n}(1+\frac{x}{n})^{-n+1}$$
for $n>1$. I think the OP can handle the rest
EDIT: Let me add some more detail. We have
$$\int_0^1\frac{dx}{(1+\frac{x}{n})^n}=\frac{n}{1-n}(1+\frac{1}{n})^{-n+1}-\frac{n}{n-1}$$
Now as $n\to \infty$, $\frac{n}{1-n}=\frac{1}{\frac1n-1}\to -1$. Since $e=\lim_{n\to \infty}(1+\frac{1}{n})^{n}$ the result follows.
A: HINT: Just evaluate the integral. For $n>1$ you have 
$$\int_0^1\frac{dx}{(1+x/n)^n}=\int_0^1\left(1+\frac{x}n\right)^{-n}dx=\left[\frac{n}{n+1}\left(1+\frac{x}n\right)^{-n+1}\right]_0^1\;;$$
evaluating that leaves you with a limit that involves pieces that ought to be pretty familiar.
A: Another way:
The sequence of unctions $\frac {1}{(1+x/n)^n}$ is dominated by the integrable function $g$ (where $g(x)=1$ for all $x\in [0,1]$, now you can use Lesbesuge dominated convergence theorem.
A: $$\int_0^1\frac{dx}{\left(1+\frac{x}{n}\right)^n}=n\int_0^1\left(1+\frac{x}{n}\right)^{-n}\left(\frac{1}{n}dx\right)=\left.\frac{n}{1-n}\left(1+\frac{x}{n}\right)^{1-n}\;\right|_0^1=$$
$$=\frac{n}{1-n}\left[\left(1+\frac{1}{n}\right)^{1-n}-1\right]=\frac{n}{1-n}\left(\left(1+\frac{1}{n}\right)\left(\left(1+\frac{1}{n}\right)^n\right)^{-1}-1\right)\xrightarrow [n\to\infty]{} (-1)\left(1\cdot e^{-1}-1\right)=1-e^{-1}$$
