I know that if the limit of the second sequence is $0$ then the limit of the product is $0$,but what if the second sequence has the limit $\infty$?

closed as unclear what you're asking by zhw., Gibbs, amWhy, T. Bongers, user10354138 Dec 7 at 2:39

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  • This needs to be restated. It's unclear what you're asking. – zhw. Dec 6 at 17:34
up vote 0 down vote accepted

Hint

$$f(x)=x\sin\bigg(\dfrac{1}{x}\bigg).$$

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