How to show that for $x\in \mathbb{R}$, $x\neq -1$, the limit $$f(x)=\lim\limits_{n\to\infty} \Big(\frac{x^n-1}{x^n+1}\Big)^2$$ exists? Determine the set of points $x\in\mathbb{R}\setminus \{-1\}$ in which $f:\mathbb{R}\setminus\{-1\}\to\mathbb{R}$ continuous.
My try: for $|x|<1$ the limit is $1$, since $x^n\to 0$ as $n\to\infty$ for such $x$.
Furthermore $f(1)=0$. It remains to consider what happens for $x\in (1,\infty)$ and $x\in (-\infty,-1)$. In both cases, $x^n\to \infty$, right? But how to proceed? L'Hôpital? Or are there other ways to do it?
And how to do the continuity part?
HINT: note that
$$\frac{x^n-1}{x^n+1}=\frac{x^n+1-1-1}{x^n+1}=1-\frac2{x^n+1}$$
and that
$$\lim_{x\to a}[f(x)]^2=[\lim_{x\to a}f(x)]^2$$
For the continuity part it is enough to see what kind of set is the image of $f$.