Prove or disprove $\lim\limits_{n→∞}\frac{x_n-1}{x_n+1} = 0\Rightarrow\lim\limits_{n→∞}x_n = 1$

Question

How to prove or disprove that $$ \lim_{n\to \infty}\frac{x_n - 1}{x_n + 1} = 0 \Longrightarrow \lim_{n\to \infty}x_n = 1 ?$$ My attempt is:$$ 0 < |x_n - 1| = \left| (x_n - 1) · \frac{x_n + 1}{x_n + 1} \right| = \left| \frac{x_n - 1}{x_n + 1} · (x_n + 1) \right|. $$ Is there any way to show that $x_n + 1$ is bounded?

Answer 1

Define $f(x)=(x-1)/(x+1)$. Suppose $f(x_n)\to0$. If $x_n$ is unbounded then there is a subsequence $x_{n_j}$ with $|x_{n_j}|\to\infty$, which implies that $f(x_{n_j})\to1$.

Answer 2

Correct me if wrong:

$$\lim_{n \to \infty} (1-\frac{2}{x_n+1})= 0$$

$$\Downarrow$$

$$\lim_{n \to \infty}\frac{2}{x_n+1}=1$$

$$\Downarrow$$

$$\lim_{n \to \infty} x_n=1.$$

Used:

1) If $(a_n +b_n)$ and $(-b_n)$ converge,

then the sum converges and

$\lim( (a_n+b_n)+ (-b_n))= \lim a_n.$

2) If $c_n$ converges and $\lim c_n \not = 0$ then

$1/c_n$ converges and

$\lim(1/c_n) = \dfrac{1}{\lim c_n}.$