Showing that $\lim\limits_{{x\to 1}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]$, $x\neq -1$ exists This problem seems too difficult for me to solve. Hence, I need the help of those in the house by providing me a detailed solution. The question actually requires me to show that if $$f(x)=\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2,\quad x\neq -1,$$ then I should find $\lim\limits_{{x\to 1}}f(x)$. 
Is it also possible to redefine $f(-1)$ in such a way that $f$ is continuous at $-1$?
 A: The inner fraction tends to $1$ for $x>1$ and to $-1$ for $0<x<1$. Then after squaring, the limit to $x\to1$ is $1$.
The behavior is quite similar around $x=-1$, despite the alternating sign. So $f(-1):=1$.
A: Noting that $x>1$ if $x\to1^+$, one has
$$\lim\limits_{{x\to 1^+}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]=\lim\limits_{{x\to 1^+}}\left[\lim\limits_{{n\to\infty}}\left(\frac{1-\frac{1}{x^n}}{1+\frac{1}{x^n}}\right)^2\right]=\lim\limits_{{x\to 1^+}}\left[\lim\limits_{{n\to\infty}}\left(\frac{1-0}{1+0}\right)^2\right]=1.$$
Also noting that $x<1$ if $x\to1^-$, one can choose $0<x<1$ and  one has
$$\lim\limits_{{x\to 1^-}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]=\lim\limits_{{x\to 1^+}}\left[\lim\limits_{{n\to\infty}}\left(\frac{0-1}{0+1}\right)^2\right]=1.$$
So
$$ \lim\limits_{{x\to 1^-}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]=\lim\limits_{{x\to 1^+}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]=1 $$
which implies
$$ \lim\limits_{{x\to 1}}\left[\lim\limits_{{n\to\infty}}\left(\frac{x^n-1}{x^n+1}\right)^2\right]=1 $$
exists.
A: What if we take $x^{2n}$ common from both numerator and denominator, cancel each out. That leads numerator to $$1-\frac{1}{x^n}$$ and denominator to $$1+\frac{1}{x^n}$$ on which, when limits are applied become, $$\frac{1 - 0}{1+0} = 1 = f(x).$$
