Find a sequence $a_n$ such that its $\lim(a_n)\rightarrow 0$ and $b_n$ that $\lim(b_n)$ has no limit (finite or infinite) such that $\lim(a_nb_n)\rightarrow1$
using arithmetic i need to find a sequence $a_n$ and $(a_n)^{-1}$.

let say $a_n$=$\frac{1}{n}$ what $b_n$ I need to choose?


Your sequence $a_n = n^{1/n}$ tends to $1$ instead of $0$ as you want.

Try the following sequence pair instead: $$a_n=1/n \text{ and } b_n=n$$

Another example:

$$a_n = (-1)^n/n \text{ and }b_n=(-1)^n n$$

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    $\begingroup$ yes I edited $a_n$ but $b_n=n$ has a limit (infinite) $\endgroup$ – gbox Dec 3 '14 at 13:42
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    $\begingroup$ @gbox $b_n$ has no limit. If you want $b_n$ not to properly diverge to $+\infty$ or $-\infty$, try the second example. $\endgroup$ – Adhvaitha Dec 3 '14 at 13:42
  • $\begingroup$ I need $b_n$ like $(-1)^n$, sorry I am new to calculus how does it called? $\endgroup$ – gbox Dec 3 '14 at 13:44
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    $\begingroup$ @gbox See the second example, I have added in my post. Usually, when you say that the limit doesn't exist, this means the sequences are allowed to properly diverge to $+\infty$ or $-\infty$. (In fact, even in your post you seem to indicate that: "...has no limit (finite or infinite) such that..." $\endgroup$ – Adhvaitha Dec 3 '14 at 13:46
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    $\begingroup$ @gbox Typically, no limit includes the case when the limit is $+\infty$ or $-\infty$. Again it depends on what convention you want to follow. $\endgroup$ – Adhvaitha Dec 3 '14 at 13:48

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