# $(x_n)$ is eventually constant if Cauchy and formulation of limit

$1)$ If $(x_n)$ is Cauchy and $x_n \in \mathbb{Z} \,\,\forall n \in \mathbb{N}$ then $x_n$ is eventually constant (for $n \geq N, x_n = x_{n+1})$

Attempt: If $(x_n)$ is Cauchy then for all $\epsilon > 0,\,\, \exists N \in \mathbb{N}$ such that $|x_n - x_m| < \epsilon$ for $n,m \geq N$.
Hence pick $\epsilon = 1$. Then we are done already since if $|x_n - x_m| < 1$, implies that the difference between two integers (absolute value) is strictly less than 1. Since the statement $|x_n - x_m| < 1$ holds for all $n,m$ take two consecutive integers. Their difference is exactly 1. So the only way for the inequality to hold is if $x_n = x_m\,\forall n,m,$so in particular $x_n = x_{n+1}$.

Is it okay?

2)Formulate a definition for the following: $\operatorname{lim}_{x \rightarrow a^+} f(x) = - \infty$

Attempt: $f(x) \rightarrow -\infty$ as $x \rightarrow a^+$ if for every $N > 0$ such that if $a < x < a + \delta$ then $f(x) < -N$

Is it okay?

Many thanks.

If $(x_{n})$ is Cauchy, then for any $\epsilon > 0$ we have $| x_{n} - x_{m}| < \epsilon$ for $n,m > N \in \mathbb{N}$. Since $(x_{n}) \in \mathbb{Z}$, $x_{n} = x_{m}$ for all $n,m > N$.
For 2), you maybe miss a quantifier: for all $N>0$ there exists $\delta$...
• Do you mean for all $N > 0,$ there exists $\delta$ such that if $a < x< x + \delta$ then $f(x) < -N$? – CAF Mar 20 '13 at 21:20