For which values of $x$ does $\frac {x+x^n}{1+x^n}$ converge and to what? Part 2 I got several good answers to this question, but they all seemed to use the algebra of limits which I'm not supposed to use on this yet.  But I've been working on it since yesterday and I think I've gotten a proof using only the definition of convergence.  Here's the exercise.

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.

Can you look at my proof below and see if it's correct or if I can improve it?  Is there some way to combine all these different cases?
Proof:
$x=1$ case:
$$s_n(1) = \frac{1+1^n}{1+1^n} = 1$$ for all $n$.  Therefore this is a constant sequence whose $n$th element is $1$ and thus it converges to $1$.
$x=0$ case:
$$s_n(0) = 0$$ for all $n$.  Similarly as above $s_n(0)$ converges to $0$.
$|x|>1$ case:
Choose $\epsilon \gt 0$ with $x\in(-\infty,-1)\cup(1,\infty)$.  Then let $N$ be some natural number grater than $\dfrac{\ln(\epsilon)}{\ln\left(\frac 1{|x|}\right)}+1$.  Then 
$$n\ge N \gt \frac{\ln(\epsilon)}{\ln\left(\frac 1{|x|}\right)}+1 \implies (n-1)\ln\left(\frac{1}{|x|}\right)\lt \ln(\epsilon) \\ \implies \left(\frac 1{|x|}\right)^{n-1} \lt \epsilon$$
and also
$$\left|\frac{x+x^n}{1+x^n}-1\right| = \left|\frac{x-1}{1+x^n}\right|\le \left(\frac{1}{|x|}\right)^{n-1}\lt \epsilon$$
Thus $s_n(x)$ for $|x|>1$ converges to $1$.
$0\lt x\lt 1$ case:
Choose $\epsilon>0$ and $x\in(0,1)$.  Then let $N$ be some natural number greater than $\dfrac{\ln\left(\dfrac{\epsilon}{2}\right)}{\ln(x)}$.  Then
$$n\ge N \gt \frac{\ln\left(\frac{\epsilon}{2}\right)}{\ln(x)} \implies 2|x|^n\lt \epsilon$$
and also
$$\left|\frac{x+x^n}{1+x^n}-x\right| = \frac{|x^n-x^{n+1}|}{|1+x^n|} \le \frac{|x|^n}{|1+x^n|} + \frac{|x|^{n+1}}{|1+x^n|} \lt 2|x|^n \lt \epsilon$$
Thus $s_n(x)$ converges to $x$ when $x\in(0,1)$.
$-1\lt x \lt 0$ case:
Choose $\epsilon>0$ and $x\in(-1,0)$.  Then let $N$ be some natural number larger than $\max\left(\frac{\ln(\epsilon/4)}{\ln(|x|)},\frac{\ln(1/2)}{\ln(|x|)}\right)$.  Then
$$n \ge N \gt \frac{\ln(\frac {\epsilon}{4})}{\ln(|x|)} \implies 4|x|^n\lt \epsilon \\ \text{and } n\ge N \gt \frac{\ln(\frac 12)}{\ln(|x|)}\implies |x|^n\lt \frac 12$$
Then because 
$$\left|\frac{x+x^n}{1+x^n}-x\right| = |1-x|\frac{|x|^n}{|1+x^n|}\lt \frac{2|x|^n}{|1+x^n|}$$
and because
$$|x|^n \lt \frac 12 \implies -\frac 12 \lt x^n \lt \frac 12 \\ \implies \frac 12 \lt 1+x^n \lt \frac 32 \\ \implies 2 \gt \frac 1{1+x^n} \gt \frac 23 \gt -2 \\ \implies \left|\frac 1{1+x^n}\right| \lt 2$$
we get
$$\left|\frac{x+x^n}{1+x^n}-x\right|\lt 4|x|^n\lt \epsilon$$
Therefore $s_n(x)$ converges to $x$ when $x\in(-1,0)$.$\square$
 A: Maybe you're still not allowed to use the algebra of limits, but you can do some transformations for getting an intuition of what you want to prove.
You've treated correctly the cases $x=0$ and $x=1$.
If $0<|x|<1$, the sequence $x^n$ converges to zero, so the task will be to prove that
$$
\lim_{n\to\infty}\frac{x+x^n}{1+x^n}=x
$$
If, instead, $|x|>1$, you can write (for $n>1$, but it's not restrictive)
$$
\frac{x+x^n}{1+x^n}=\frac{\dfrac{1}{\mathstrut x^{n-1}}+1}{\dfrac{1}{x^n}+1}
$$
and note that $1/x^n$ converges to zero, so the task is to prove that
$$
\lim_{n\to\infty}\frac{x+x^n}{1+x^n}=1
$$
Case $0<|x|<1$
Let $\varepsilon>0$ and consider the inequality
$$
\left|\frac{x+x^n}{1+x^n}-x\right|<\varepsilon
$$
that becomes
$$
\left|\frac{x^n}{1+x^n}(1-x)\right|<\varepsilon
$$
Take $N_0$ so that $|x^n|<1/2$ for $n>N_0$. Then
$$
\frac{1}{2}<1+x^n<\frac{3}{2}
$$
and, in particular, $1/(1+x^n)<2$. Thus the task is to find $N_1$ such that, for $n>N_1$, $|x^n|<\frac{\varepsilon}{2(1-x)}$.
Indeed, if we find $N_1$ and take $n>\max(N_0,N_1)$, then
$$
\left|\frac{x^n}{1+x^n}(1-x)\right|<
|x^n|\cdot 2(1-x)<\frac{\varepsilon}{2(1-x)}2(1-x)=\varepsilon
$$
Thus, both $N_0$ and $N_1$ are provided once we show $\lim_{x\to\infty}x^n=0$, for $0<|x|<1$ (proof at end).
Case $|x|>1$
Let $\varepsilon>0$ and set $x=1/t$. The inequality
$$
\left|\frac{x+x^n}{1+x^n}-1\right|<\varepsilon
$$
becomes
$$
\left|\frac{t^{n-1}+1}{t^n+1}-1\right|<\varepsilon
$$
or
$$
\left|\frac{t^{n-1}(1-t)}{1+t^n}\right|<\varepsilon
$$
This is essentially the same as before, because $0<|t|<1$.
Proof of $|x^n|\to0$ for $0<|x|<1$.
Let $1/|x|=1+t$, so $t>0$. By Bernoulli's inequality, $(1+t)^n>1+nt$ and so
$$
|x|^n<\frac{1}{1+nt}
$$
Thus, in order to get $|x^n|<\varepsilon$, we just need to have
$$
\frac{1}{1+nt}<\varepsilon
$$
that is
$$
n>\frac{1}{t}\left(\frac{1}{\varepsilon}-1\right)
$$
