Suppose $<a_n>$ and $<b_n>$ are sequences with $a_n$, $b_n$ $>0$ for all $n$, and there is a real number $l>0$ such that $a_n/b_n \rightarrow l$ as $n \rightarrow \infty$. Then show: There exists a number $N>0$ such that ${l/2}*b_n<a_n<(3l/2)*b_n$ for all $n>N$.
Since we know that $a_n/b_n \rightarrow l$ as $n \rightarrow \infty$, then for $\varepsilon>0$ $\exists N\in R$, $\forall n>N$ s.t. $|a_n/b_n -l|<\varepsilon$
Since $b_n$ is positive, then: $|a_n -l*b_n|<\varepsilon*b_n$
Now, by considering left hand side: $|a_n -l*b_n| = |a_n - 3/2l*b_n + 1/2l*b_n| \leq |a_n - 3/2 l*b_n| + |1/2l*b_n|$ by using triangular inequality.
From here I do not know what to do. Perhaps I did not even need to do what I just wrote but those were my thoughts.
Thanks a million!
WLOG, we can assume $l=1$.
take $$\epsilon=\frac{1}{2}.$$
$$\lim_{n\to+\infty}\frac{a_n}{b_n}=1 \implies$$
$\exists N\geq0 : \forall n>N $
$$1-\epsilon<\frac{a_n}{b_n}<1+\epsilon\implies$$
$$(\forall n>N)\;\;\frac{1}{2}b_n<a_n<\frac{3}{2}b_n.$$
if $l\neq1$ replace $b_n$ by $lb_n$