If $\lim (s_n – s)/(s_n + s) = 0$, prove that $\lim s_n = s$. I would like to prove that if $\lim (s_n – s)/(s_n + s) = 0$, then $\lim s_n = s$.
I try $|s_n - s| = |s_n + s |\frac{|s_n - s|}{|s_n +s|}$. 
So $|s_n - s| < |s_n + s|\epsilon$. But I can't show that $s_n$ is bounded. So I can't complete the proof. 
 A: $$\lim_{n\to\infty}\frac{s_n-s}{s_n+s}=\lim_{n\to\infty}\frac{s_n+s}{s_n+s}-\frac{2s}{s_n+s}=\lim_{n\to\infty}\frac{s_n+s}{s_n+s}+\lim_{n\to\infty}\frac{-2s}{s_n+s}=1-\lim_{n\to\infty}\frac{2s}{s_n+s}=0$$
$$\lim_{n\to\infty}\frac{1}{s_n+s}=\frac{1}{2s}$$
$s\ne 0$, because otherwise the condition given cannot be true. Hence, the above limit exists.
$$\frac{\lim_{n\to\infty}1}{\lim_{n\to\infty}s_n+s}=\frac{1}{2s}$$
$$\lim_{n\to\infty}(s_n+s)=2s$$
$$\lim_{n\to\infty}s_n=s$$
A: To show that $s_n$ is bounded, note that $$\left|\frac{s_n-s}{s_n+s}\right|=\left|1-\frac{2s}{s_n+s}\right|\geq1-\frac{|s|}{||s_n|-|s||}\geq\frac12$$ if $|s_n|\geq3|s|$. By the convergence of $\frac{s_n-s}{s_n+s}$, we must have $|s_n|\leq3|s|$ for $n$ large enough.

Alternatively,
$$\begin{align*}&\lim_{n\to\infty}\frac{s_n-s}{s_n+s}=0\\
\iff&\lim_{n\to\infty}\left(1-\frac{2s}{s_n+s}\right)=0\\
\iff&\lim_{n\to\infty}\frac{s}{s_n+s}=\frac12\\
\iff&s\neq0\text{ and }\lim_{n\to\infty}\frac1{s_n+s}=\frac1{2s}\\
\iff&s\neq0\text{ and }\lim_{n\to\infty}(s_n+s)=2s\\
\iff&s\neq0\text{ and }\lim_{n\to\infty}s_n=s\end{align*}$$
A: If $s_n$ is not bounded then when you take the limit, $\lim_{n \to \infty} \frac{s_n - s}{s_n + s}$ gives you $1$, not $0$ by L'Hôspital, doesn't it?
