Different proofs of $\lim\limits_{n \rightarrow \infty} n \int_0^1 \frac{x^n - (1-x)^n}{2x-1} \mathrm dx= 2$ It can be shown that
$$ n \int_0^1 \frac{x^n - (1-x)^n}{2x-1} \mathrm dx = \sum_{k=0}^{n-1} {n-1 \choose k}^{-1}$$
(For instance see my answer here.)
It can also be shown that $$\lim_{n \to \infty} \ \sum_{k=0}^{n-1} {n-1 \choose k}^{-1} = 2$$
(For instance see Qiaochu's answer here.)
Combining those two shows that
$$ \lim_{n \to \infty} \ n \int_0^1 \frac{x^n - (1-x)^n}{2x-1} \mathrm dx  = 2$$
Is there a different, (preferably analytic) proof of this fact? Please do feel free to add a proof which is not analytic.

Let $x=(1+s)/2$, so that the expression becomes
$$ \frac{n}{2^{n+1}} \int_{-1}^1 \frac{(1+s)^n-(1-s)^n}{s} ds =
\frac{n}{2^{n}} \int_{0}^1 \frac{(1+s)^n-(1-s)^n}{s} ds. $$
(The integrand is an even function.)
Fix some $c$ between 0 and 1, say $c=1-\epsilon$.
Then the integral from 0 to $c$ will be small in comparison to $2^n$,
since the integrand is bounded by a constant times $(1+c)^n$,
so as far as the limit is concerned it is enough to look at the
integral from $c$ to 1,
and in that integral we can neglect the term $(1-s)^n$ since it will
also contribute something much smaller than $2^n$.
The surviving contribution therefore comes from
$$ \int_c^1 \frac{(1+s)^n}{s} ds, $$
which lies between
$$ \int_c^1 \frac{(1+s)^n}{1} ds $$
and
$$ \int_c^1 \frac{(1+s)^n}{c} ds, $$
that is,
$$ \frac{2^{n+1} - (1+c)^{n+1}}{n+1} < \int_c^1 \frac{(1+s)^n}{s} ds < \frac{1}{c} \frac{2^{n+1} - (1+c)^{n+1}}{n+1} .$$
Multiplying by $n/2^n$ and letting $n\to\infty$ shows that the liminf
is at least 2 and the limsup is at most $2/c$.
But since this holds for any $c$ between 0 and 1,
it follows that liminf=limsup=2, hence the limit is 2.
