Cauchy sequence convergence Suppose that $a_n$ and $b_n$ are Cauchy sequences, and that $a_n &lt; b_n$ for all n. Prove that $\lim_{x \to \infty}a_n \le \lim_{x \to \infty}b_n$ for all n.

Is it sufficient to say that we know both Cauchy sequences must converge to the limit, and since $a_n$ is always less than $b_n$, the limits will follow the desired inequality?

Edit: since this is not true, what would be the appropriate strategy to prove?
Let $a_n$ and $b_n$ be your two sequences such that $a_n&lt;b_n$ for all $n$. Consider the
sequence $\{b_n-a_n\}$ which is greater than $0$ for all $n$. This sequence can also be seen to be Cauchy. Of course, all Cauchy sequences in $\mathbb{R}$ converge. Call its limit $x$. Suppose its limit was strictly
less than $0$. Since $x$ is less than $0$, there is a $\varepsilon&gt;0$ such that
$(x-\varepsilon,x+\varepsilon)\subseteq (-\infty,0)$. Now since $\lim_{n\rightarrow\infty} \{b_n-a_n\}=x$, there exists an $N$ such that $b_N-a_N\in(x-\varepsilon,x+\varepsilon)$; that is, $b_N-a_N&lt;0$ which is a contradiction. We must have the limit is at least $0$. In the case
that both limits are finite, it's an easy lemma that $\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ (which is our case as both $a_n$ and $b_n$ are Cauchy sequences). Thus $0\leq\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ which implies $\lim_{n\rightarrow\infty} a_n \leq \lim_{n\rightarrow\infty} b_n$.
It is not true, take $b_n=1/n=-a_n$, the strict inequality is not respected at the limit.