Prove that if a real convergent sequence with limXn=x as n--> infinity, if $X_n<6$, then $x\le6$ and if $X_n>2$ then $x\ge2$

Question

I am very new to this and I am finding it very difficult to do proofs, here is my attempt.

If for each $n\in\mathbb N$ $X_n<6$, prove that $x≤6$

Let $\varepsilon > 0$ be given. since $(Xn)$ converges

$|X_n-x|<\varepsilon$ and we are told $X_n<6$

$|x|= |x-X_n+X_n| \le |X_n-x| +|X_n|$

$|x| < \varepsilon + |Xn|$

$|x| < \varepsilon + 6$

implies $ x\le6 $

If for each $n\in\mathbb N$, $X_n>2$, then prove that $x\ge2$

Let $\varepsilon > 0$ be given. since $(X_n)$ converges and we are told $X_n>2$

$|X_n| = |X_n-x+x|\le|X_n-x| +|x|$

$|X_n| < \varepsilon + |x| $

$|X_n|-\varepsilon < |x|$

$2-\varepsilon < |x|$

implies $|x|\ge 2$

Any feed back most welcome and many thanks in advance

2nd attempt at 2nd part of question

If for each $n\in\mathbb N$, $X_n>2$, then prove that $x\ge2$

Let $\varepsilon > 0$ be given. since $(X_n)$ converges and we are told $X_n>2$

$|X_n-x|< \varepsilon $

$ -\varepsilon < X_n-x< \varepsilon$

$ -x < \varepsilon -X_n$

$ x > 2 - \varepsilon $

implies $ x\ge 2$

$2-\varepsilon < |x|$ doesn't imply $|x| \ge 2$ on its own, it only implies $|x|>2-\varepsilon$ (use the fact that it holds for any $\varepsilon>0$). More importantly, $|x| \ge 2$ doesn't imply $x \ge 2$ (from only this, $x$ can be $-3$). Use $|X_n-x|<\varepsilon \implies x < X_n+\varepsilon,, x > X_n-\varepsilon$. — Abstraction, Dec 17 '15 at 15:30

score 0 · Answer 1 · answered Dec 17 '15 at 15:36

0

Alternative proof:

Assume that $x>6$. Then $\epsilon:=6-x>0$.

If $n$ large enough then $x_n\in(x-\epsilon,x+\epsilon)\subset(6,\infty)$.

This contradicts $x_n\leq6$, so the assumption is wrong.

answered Dec 17 '15 at 15:36

drhab

1 Answers1