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We need to proof that $b_n/a_n\rightarrow0$ for $a_n\rightarrow\pm\infty$ and $(b_n)$ is restricted. But I came to $|b_n| \cdot 1/|a_n| \lt \epsilon$ and now I'm really stuck.. Can someone help me?

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hint: Let us suppose for the moment that $a_n\to+\infty$ Use the squeezing theorem. Since $(b_n)$ is bounded, there exists a costant $M>0$ such that $-M<b_n<M$, for every $n$ but then $-M/a_n<b_n/a_n<M/a_n$... For the case $a_n\to-\infty$ there is a change in the sign of the inequalities, but not much more than that. Can you go on from here? –  uforoboa Sep 23 '11 at 10:02
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Let $\varepsilon > 0$. Since $(b_n)$ is bounded there exists a constant $M$ such that $|b_n| < M$ for all $n$. Moreover $|a_n| \to \infty$ hence for some $N$ we get $|a_n| > M/\varepsilon$ for $n > N$. It is equivalent to $|1/a_n| < \varepsilon/M$.

Finally we obtain $$\Bigg|\frac{b_n}{a_n}\Bigg| < \varepsilon$$ for $n > N$ which means that $b_n/a_n \to 0$.

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Thanks, but why $|a_n|>M/\epsilon$? I thought $n$ depends on $\epsilon$, so when you take $n$ large enough, $\epsilon$ will become smaller and smaller? –  Kevin Sep 23 '11 at 10:19
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@Kevin: the order is the following. 1. You fix $\varepsilon$, 2. you choose $N'(\varepsilon)$ large enough to ensure the inequality $|a_n|>M/\varepsilon$ for all $n>N'(\varepsilon)$. With increase of $n$, the value of $\varepsilon$ stays unchanged since you fixed it from the beginning. We are only interested that the inequality holds. –  Ilya Sep 23 '11 at 10:25
    
@xen: why do you need $N$? If $b$ is bounded, than there is such $M$ that $|b_n|<M$ for all $n$? You're correct, of course, but it may be just a bit confusing for a person who only started to learn limits. –  Ilya Sep 23 '11 at 10:27
    
Thanks, I get it now :-)! –  Kevin Sep 23 '11 at 10:39
    
@Gortaur: Oh, of course. Thanks. –  xen Sep 23 '11 at 17:21
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