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I'm studying for an exam and I'm not too sure how to do this problem.

Let $a_n$ be a sequence in $\mathbb R$ with $\lim_{n \rightarrow \infty}a_n = -1$. Prove that there exists $n_0 \in \mathbb N$ such that $a_n < -\frac{1}{2}$ for all $n \ge n_0$

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  • $\begingroup$ what is your definition of limit ? $\endgroup$
    – user87543
    Dec 10, 2013 at 12:39

1 Answer 1

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$a_n\to -1<=>\forall ε>0$ there is a $n_0:|a_n+1|<ε$ for a every $n\geq n_0$.This means that $-1-ε<a_n<ε-1$ for every $n\geq n_0$. Take $ε=\frac {1}{2}$.

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  • $\begingroup$ please... do not do this.. give OP some hint and some time... you may want to rewrite this as "given $\epsilon >0$ instead of $\forall \epsilon >0$" $\endgroup$
    – user87543
    Dec 10, 2013 at 12:43
  • $\begingroup$ How would I proceed with ε = $1\over2$? I got all of what you put before on my previous exam but I didn't know how to prove it. $\endgroup$
    – Brian
    Dec 10, 2013 at 12:44
  • $\begingroup$ @Brian, because this stands for every $ε>0$,then it stands for $ε=\frac {1}{2}$ too. So if you put $ε=\frac {1}{2} $ in the double inequality ,you will have that $a_n<-\frac {1}{2}$ for every $n\geq n_0$. $\endgroup$
    – Haha
    Dec 10, 2013 at 12:46
  • $\begingroup$ OH, sorry I'm running low on sleep. Thanks! I got it now $\endgroup$
    – Brian
    Dec 10, 2013 at 12:47

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