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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?

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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

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

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  • $\begingroup$ In general the existence of $\lim_{n\to \infty}(a_n+b_n)$ does not imply the existence of the limits $\lim_{n\to \infty} a_n$ and $\lim_{n\to \infty} b_n$. But in this case where $a_n$ is a constant sequence I guess it will work. $\endgroup$ – PtF Jun 24 '18 at 15:04
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    $\begingroup$ @PtF $\lim (a_n + b_n)$ existing along with $\lim b_n$ existing imply that $\lim (a_n + b_n) - \lim b_n = \lim (a_n +b_n -b_n) = \lim a_n$. $\endgroup$ – fleablood Jun 24 '18 at 15:09
  • $\begingroup$ PtF. Lim (1-1/(x_n+1)) exists lim (-1) exists , add to get lim( (1-1/x_n+1))+ lim (-1) = lim( (1-...) -1).is this ok? $\endgroup$ – Peter Szilas Jun 24 '18 at 15:12
  • $\begingroup$ .... and if one of the "sequences" is constant it's .... even more basis. If $\lim a_n = k$ then $\lim (a_n + m) = k+m$ and $\lim m*a_n = mk$ and if $k \ne 0$ then $\lim \frac m{a_n} = \frac mk$. That is a very basic result. $\endgroup$ – fleablood Jun 24 '18 at 15:14
  • $\begingroup$ +1 for the simple argument. Laws of limits are actually more powerful than what most beginners think. $\endgroup$ – Paramanand Singh Jun 25 '18 at 3:02
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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 $f(x_{n_j})\to1$.

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