Q:

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.

