Limit of exponentials Why is $n^n (n+m)^{-{\left(n+m\over 2\right)}}(n-m)^{-{\left(n-m\over 2\right)}}$ asymptotically equal to $\exp\left(-{m^2\over 2n}\right)$ as $n,m\to \infty$?

Let $m = n x$.
Take the logarithm:
$$
  n \log(n) - n \frac{1+x}{2} \left(\log n + \log\left(1+x\right) \right) - n \frac{1-x}{2} \left( \log n + \log\left(1-x\right) \right) 
$$
Notice that all the terms with $\log(n)$ cancel out, so we are left with
$$
   -\frac{n}{2} \left( (1+x) \log(1+x) - (1-x) \log(1-x) \right) 
$$
It seems like the you need to assume that $x$ is small here, meaning that $ m \ll n$. Then, using Taylor series expansion of the logarithm:
$$
  (1+x) \log(1+x) + (1-x) \log(1-x) = (1+x) \left( x - \frac{x^2}{2} + \mathcal{o}(x^2) \right) + (1-x) \left(-x - \frac{x^2}{2} + \mathcal{o}(x^2)\right) = x^2 + \mathcal{o}(x^3)
$$ 
Hence the original expression, asymptotically, equals
$$
    \exp\left( -\frac{n}{2} x^2 + \mathcal{o}(n x^3)\right) = \exp\left(- \frac{m^2}{2n} + \mathcal{o}\left(\frac{m^3}{n^2}\right) \right)
$$
