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I have to use the definition of convergence only to prove this. This is my attempt on proving this statement:

Let $\epsilon >0$. Then, we know there is some $N\in\mathbb{N}$ such that for all $n \geq N$ we have $|x_n-L| < \epsilon$. Thus,

$$|x_n-L|<\epsilon$$ $$\Leftrightarrow L-\epsilon<x_n<L+\epsilon$$ Since $a \leq x_n \leq b$, we can deduce that $$L-\epsilon<b \implies > L<b+ \epsilon$$ $$a < L+\epsilon \implies a-\epsilon < L$$ And $a-\epsilon <L < b+\epsilon$. $\blacksquare$

This is as far as I can get but does the last statement imply that $a\leq L \leq b$? Since we can make $\epsilon$ as small as we want, $a-\epsilon <L < b+\epsilon$ will eventually becomes $a < L <b$ but it doesn't necessarily mean $a\leq L \leq b$.

EDIT: Also, how can I place my black box (QED symbol) to the corner?

user3000482
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  • You're right. More formally, since $\epsilon$ was arbitrary, this is true for all $\epsilon > 0$. Now argue by contradiction: if $a>L$, then for $\epsilon = a-L$, we have $a - (a - L) < L$, i.e. $0<0$. Same applies for $b$. –  Jan 29 '17 at 19:50
  • If $L$ were $a$ (or $b$), then it would satisfy $a-\epsilon<L<b+\epsilon$ for every $\epsilon>0$, but it would not satisfy $a<L<b$. In other words, when you drop $\epsilon$ to 0, the statement you should end up with is $a\leq L\leq b$, and not $a<L<b$. – RideTheWavelet Jan 29 '17 at 19:52
  • @RideTheWavelet What about for the possibility of $b<L<b+\epsilon$? Since we don't know yet if $a \leq L \leq b$, isn't it possible for $L>b$ and $L<a$? – user3000482 Jan 29 '17 at 19:56
  • @OpenBall That is much better proof! Thank you. – user3000482 Jan 29 '17 at 19:57
  • Please don't use proof-verification tag as the only tag on your question. Use other tags to indicate what area of mathematics the proof comes from. – Michael Albanese Feb 01 '17 at 14:55
  • @MichaelAlbanese Sorry, I will keep that in mind for my next questions! – user3000482 Feb 01 '17 at 18:37

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