For which values of $x$ does $\frac {x+x^n}{1+x^n}$ converge and to what? 
Let $$s_n(x) = \frac{x+x^n}{1+x^n}$$ for all $n\in\Bbb N$ and $x\in \Bbb R\setminus\{-1\}$.  Find each real number $x$ for which the sequence $(s_n)$ is convergent and find the limit of the sequence.

I've broken it into cases.  Here's what I've been able to get so far.
$x\in[1,\infty)$ case: $$\left|\frac{x+x^n}{1+x^n}-1\right| = \left|\frac{x+x^n-(1+x^n)}{1+x^n}\right| = \left|\frac{x-1}{1+x^n}\right|\lt \frac{x}{x^n} = \left(\frac 1{x}\right)^{n-1}$$
which gets arbitrarily small as $n\to \infty$ (in fact it's easy to show that $1/{x^n} \to 0$), so we should be able to bound this by any $\epsilon$ for a large enough value of $N$. (Does this argument work?)
$x\in(-1,1)$ case:
$$\left|\frac{x+x^n}{1+x^n}-x\right| = \left|\frac{x+x^n-x-x^{n+1}}{1+x^n}\right| = \left|\frac{x^n(1-x)}{1+x^n}\right|\lt \frac{x^n}{x^n}= 1$$
At this point I'm stuck because this doesn't say anything about whether we could bound this by some $\epsilon \lt 1$.
$x\in(-\infty,-1)$ case:
I think it diverges, but I'm not sure at all on this part.

Any hints as to how to proceed from here?
Edit: Whoops.  $x=1$ should have been it's own case.  But that's the sequence whose $n$th element is $$\frac{1+1^n}{1+1^n} = 1$$ and I've already proven that constant sequences converge.  So that one's easy.
Edit 2: This exercise in from Little, Teo, and Brunt's Real Analysis via Sequences and Series in the section right before the arithmetic/ algebra of limits.  All I'm supposed to use for these is the definition of convergence of sequences.
 A: "I think it diverges, but I'm not sure at all on this part." No. In the first part, you do not really use that $x$ is positive that much, and you could come up with such a proof for that case too. But it might be easier to just do it like this. 
$$\frac{x+x^n}{1+x^n}= \frac{x^{-(n-1)}+1}{x^{-n}+1}$$
since for $|x|>1$ we have that $x^{-(n-1)}$ and $x^{-n}$ tend to $0$ as $n$ grows the limit is $1$.  

For $|x|< 1$, the same thing works but it is rather simpler. 
We have that $x^{n}$ tend to $0$ as $n$ grows the limit is thus equal to $\frac{x}{1}$. 
Finally for $x=1$ the limit is of course $1$. 
A: Why not using arithmetic of limits in some cases?:
$$x\in(-1,1)\implies \frac{x+x^n}{1+x^n}\xrightarrow[n\to\infty]{}\frac{x+0}{1+0}=x$$
$$x\in[1,\infty)\implies\frac{x+x^n}{1+x^n}\le\frac{x+x^n}{x^n}=\left(x^{-n+1}+1\right)\xrightarrow[n\to\infty]{}0+1=1$$
and etc.
A: $\frac{x+x^n}{1+x^n}=1+\frac{x-1}{1+x^n}$
$x=1\,$ => $\,1$
$x=-1\,$ => $\,1+x^n\in\{0;2\}$ alternating , therefore divergence (and $\frac{1}{0}$ is not defined)
$|x|>1\,$ => $\,\frac{x-1}{1+x^n}\to 0\,$ => $\,1$
$|x|<1\,$ => $\,x^n\to 0\,$ => $\,x$
A: In your middle case,  use the squeeze theorem,  0 times bounded is 0.  $x^n$ goes to 0, and the other part is bounded, so it goes to 0,  thus your limit is $x$
