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Prove that if $ \lim_{n \to \infty} a_n = 0 $ and $ \left \{ b_n \right \} $ is bounded, then $ \lim_{n \to\infty} (a_n b_n) = 0. $

$\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=0$ as desired.

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Oregon State University

University of Michigan - Ann Arbor

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Idaho State University

so suppose that the limit of a and zero and that being is abounded. Sequels. So this means that there exist, let's say, a positive M David than zero such that, of course here and could even be it could be zero, not a big deal. Absolute value being is always less than or equal to him. So am is a an upper bound for B and positive number here. So here I am is not equal to infinity. Yeah. Now let's use the fact that the limit of a and zero So here, let's just consider any positive number. Absolutely then, since the limit of a and zero there exist and end such that if we take a little and bigger than this big end so such that if we take a little and bigger than begin then a a. And in the absolute value, they're the distance between a and zero is less than Absalon divided by him. So this is just using the definition of limited a zero. So now, if little and is bigger than end, then luscious multiply both sides of this Ibn. So we have an bien minus zero equals just and being an absolute value. This is just a n times being now. This is less than epsilon over m times M. And the reason for this is because on one hand we know that a n an absolute value is less than epsilon over him. On the other hand, we know that BN is less than or equal to him. And if we multiply these out, the EMS cancel and we just haven't Absalon here. And this proves that's a limited of Anne times being a zero by definition, and that's our final answer.