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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 L

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

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

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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 $b$. – 2017-01-29
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    If $L$ were $a$ (or $b$), then it would satisfy $a-\epsilon0$, but it would not satisfy $a$\epsilon$ to 0, the statement you should end up with is $a\leq L\leq b$, and not $a – 2017-01-29
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    @RideTheWavelet What about for the possibility of $b$a \leq L \leq b$, isn't it possible for $L>b$ and $L – 2017-01-29
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    @OpenBall That is much better proof! Thank you. – 2017-01-29
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    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. – 2017-02-01
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    @MichaelAlbanese Sorry, I will keep that in mind for my next questions! – 2017-02-01

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