# Proving the sandwich theorem for a sequence also applies to its limit

Let $\{x_n\}$ be a sequence such that $a\leq x_n\leq b$. Suppose the sequence converges to $x$. Show that $a\leq x\leq b$.

My attempt:

Take $\epsilon>0$.

There is an $N_0$ such that $a\leq x_n\leq b$ when $n\geq N_1$.

There is an $N_1$ such that $a< \epsilon$ when $n\geq N_1$.

There is an $N_2$ such that $b<\epsilon$ when $n\geq N_2$.

Selecting $N=\max(N_0,N_1,N_2)$. Then when $n\geq N$, $-\epsilon<a\leq x_n-x\leq b < \epsilon$, and $a\leq x\leq b$.

Is this the right approach?

• No. You cannot use $a<\epsilon$, $b<\epsilon$. Youmight have $a=1$, $b=2$, for example, but $\epsilon$ is small – Hagen von Eitzen Sep 29 '16 at 6:00
• Please. Look at my edits to the question. Where in the world did you get your ideas of how to edit MathJax code? $\qquad$ – Michael Hardy Sep 29 '16 at 6:17
• @MichaelHardy. I appreciate your edit. Today is my first time using it. – ozarka Sep 29 '16 at 6:19

By definition of limit for all $\epsilon>0$ there is $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, $$x-\epsilon<x_n< x+\epsilon.$$ Now assume that $x>b$ and let $\epsilon=(x-b)/2>0$, then $$b<\frac{x+b}{2}=x-\frac{x-b}{2}<x_n\leq b$$ which is a contradiction.
What happens when $x<a$?
• Unfortunately I can't see it. I'm assuming that I need to show the inequality will not hold when $x>b$? All I've done so far is substitute $\epsilon$ with your formulation. I've just been stuck, staring at it. – ozarka Sep 29 '16 at 6:22
• I can say that: $x$ - $(x-b)$/$2$ $<$ $x$ $+$ $(x-b)$/2 to arrive at $-2x$<$-2b$ which implies $x$ $<$ $b$ but we assumed the opposite. So $b$ must be greater than $x$. Now should I assume by contradiction that $x$<$a$ and $\epsilon$ = $-$$(x-a)/2 > 0? – ozarka Sep 29 '16 at 6:29 • @ozarka No. You have that b<x-(x-b)/2<x_n\leq b. Try the case a>x. – Robert Z Sep 29 '16 at 6:51 • so I assumed that x<a. and let \epsilon = -$$(x-a)$/$2$>$0$. Then $a$ > $(x+a)/2$ = $x$ + $-$$(x-a)$/$2$ < $x_n$ $\geq$ $a$. A contradiction. ? – ozarka Sep 29 '16 at 6:55