What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$? What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$
I'm not sure which to believe - Mathematica or L'Hopital.

The interesting form is
$\left(\frac{n-x}{n+x}\right)^{n}$.
The $n^2$ just blows things up.
Taking the log,
$n \ln \frac{n-x}{n+x}
= n \ln \frac{1-x/n}{1+x/n}
= n \big(\ln (1-x/n)- \ln(1+x/n)\big)
$.
Using $\ln(1+z) = z-z^2/2+z^3/3-z^4/4+...$
and
$\ln(1-z) = -z-z^2/2-z^3/3-z^4/4+...$,
and writing $z$ for $x/n$,
this is
$n((-z-z^2/2-z^3/3-z^4/4+...)-(z-z^2/2+z^3/3-z^4/4+...))
= n(-2z-2z^3/3-...)
=-2n(x/n+x^3/(3n^3))
=-2x-2x^3/(3n^2)-...
$.
So $\left(\frac{n-x}{n+x}\right)^{n}
=\exp(-2x-2x^3/(3n^2)-...)
=\exp(-2x)exp( -2x^3/(3n^2)-...)
=\exp(-2x)(1-2x^3/(3n^2) + O(1/n^4))
$.
If you use the $n^2$,
this behaves like
$\exp(-2xn)$
which $\to 0$ if $x > 0$
and $\to \infty$ if $x < 0$.
