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I'm studying limits of a sequence and I'm confused as to the algebraic manipulation. My book defines an arbitrary sequence $a_n$ with $\lim a_n=s$. Then considers the definition of the limit; there exists an $N$ such that for $n>N$ $$\lvert a_n-s\rvert<\epsilon$$ then without explanation $$s-\epsilon<a_n$$ Where does this come from? How can this be said? How can the absolute value bars be dropped without knowing $a_n$ and $s$? I know intuitively its obvious but can it be shown directly?

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    $\begingroup$ Using $|a| < b \Leftrightarrow -b < a < b$. $\endgroup$ – user333870 May 28 '16 at 21:15
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If $\varepsilon > 0$, $$|a_n - s| < \varepsilon \iff -\varepsilon < a_n - s < \varepsilon$$

You can attack this in cases: If $a_n - s \geq 0$, then $-\varepsilon < -(a_n-s) \leq 0 \leq a_n - s < \varepsilon$ and with appropriate modifications if $a_n - s \leq 0$.

Then, adding $s$ throughout $$ s - \varepsilon < a_n < s + \varepsilon \text{.} $$

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You can do it directly: If $\epsilon > 0$ then

$$|a_n-s|<\epsilon \Leftrightarrow (a_n-s)<\epsilon, -(a_n-s)<\epsilon$$

That is,

$$-\epsilon<a_n-s<\epsilon$$

Adding $s$ in the inequality

$$s-\epsilon<a_n<s+\epsilon$$

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