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.
 A: I think this goes to zero for positive $x$.  Consider the log of the expression:
$$n^2 \log{\left(\frac{1-(x/n)}{1+(x/n)}\right)} = n^2 \left[ \log{\left(1-\frac{x}{n}\right)}-\log{\left(1+\frac{x}{n}\right)}\right]$$
For fixed $x>0$, the value of $x/n$ is small compared with $1$, so Taylor expand and get
$$n^2 \left ( -\frac{x}{n} - \frac{x}{n}\right) = -2 n x$$
which clearly $\to -\infty$ as $n \to \infty$.  Taking exponentials, $e^{-\infty} = 0$, loosely speaking.
For $x<0$, this diverges.  For $x=0$, the limit is $1$.
A: Both, and more.
If $x=0$ the limit is trivially 1. Otherwise, if $x>0$
$$
\left(\frac{n-x}{n+x}\right)^{n^2} = \left(\frac{n+x-2x}{n+x}\right)^{n^2} = \left(1 - \frac{2x}{n+x}\right)^{n^2} = \left(1 - \frac{1}{n/\left(2x\right)+1/2}\right)^{n^2} = \left(1 - \frac{1}{\bar n}\right)^{(\bar n-1/2)^24x^2} = \left(1 - \frac{1}{\bar n}\right)^{(\bar n^2 -\bar n +1/4)4x^2}
$$
Splitting the final term into a product (on the basis that $a^{b+c+d} = a^ba^ca^d$) the limit $n\to\infty \iff \bar n \to \infty$ is $0\cdot$(something bounded) $= 0$. Another way of seeing this is considering the exponent to be $\Theta(n^2)$.
Finally, if $x<0$ the limit is $\infty$ since the term in the brackets is $(1+\frac 1 {\bar n})$ when the limit is $\bar n \to \infty$.
The basis of these observations is the limit $1<e = \lim_{n\to\infty}(1+\frac 1 n)^n$, and derivations thereof, like $(1-\frac 1n)^n\to\frac 1 e<1$, etc.
A: You want $\lim f(n)$. Take logs and simplify:
$$
\ln f(n) = n^2 \ln \left(\frac{n-x}{n+x}\right)
         = n^2 \ln \left(1 - \frac{2x}{n+x}\right)
$$
and use the Taylor series
$$
\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots
$$
to get
$$
\ln f(n) = -n^2 \left( \frac{2x}{n+x}
                      + \frac{1}{2} \left(\frac{2x}{n+x}\right)^2\ldots \right)
         \to -\infty
$$
and so $f(n) \to 0$.
A: 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 $x < 0$.
A: If $x=0$ the limit is clearly $1$. 
Let $x$ be positive, and note that our expression is equal to 
$$\left(\frac{(1-\frac{x}{n})^n}{(1+\frac{x}{n})^n}   \right)^n.$$
The thing inside the large brackets has limit $e^{-2x}$. So for large enough $n$, it is less than, for example, $e^{-x}$. Thus for large enough $n$, the whole thing is $\lt e^{-nx}$. It follows that our limit is $0$.
The argument for negative $x$ is similar. Let $y=-x$. The thing inside the large brackets has limit $e^{2y}$, so for large enough $n$ it is bigger than $e^y$. Thus the whole thing is bigger than $e^{ny}$, and therefore the limit doesn't exist, or, if one prefers, is $\infty$.
A: Hint:
$L=\lim_{n\to \infty} \left(\frac{n-x}{n+x}\right)^{n^2}=\lim_{n\to \infty}\left[\left(1-\frac{2x}{n+x}\right)^{-\frac{n+x}{2x}}\right]^{-\frac{2xn^2}{n+x}}$
if a is real positive then $L=+\infty$, but if a is real negtive then $L=0$
