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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, for every $n$ but then -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?2011-09-23

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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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    @Gortaur: Oh, of course. Thanks.2011-09-23