Sequences that tend to zero If the limit of one sequence $\{a_n\}$ is zero and the limit of another sequence $\{b_n\}$ is also zero does that mean that $\displaystyle\lim_{n\to\infty}(a_n/b_n) = 1$?
 A: Given $b_n$ whose limit is 0, we can choose $a_n$ to make the limit:


*

*1: set $a_n = b_n$

*0: set $a_n = b_n^2$, or even $a_n = 0$

*Some other constant c: set $a_n = cb_n$

*Undefined: $a_n = (-1)^nb_n$

*Infinite: $a_n = \sqrt{|b_n|}$
So we can make it basically anything.
A: No. Let $a_n = \dfrac 1 n$ and $b_n = \dfrac 1 {n^2}$.
A: Of course not.


*

*$a_n=\dfrac{(-1)^n}{n}$,

*$b_n=\dfrac{1}{n}$, 

*Hence, $r_n=\dfrac{a_n}{b_n}=(-1)^n$ and the limit of $r_n$ does not even exist.

A: If we want an example such that the limit exists: Let $r \in \Bbb{R}$ be your favorite number, suppose that $b_n \to 0$, and let $a_n = r b_n$. Then $a_n \to 0$ and $a_n/b_n \to r$. 
If we want an example such that the limit does not exist: Let $b_n \to 0$ and $\{c_n\}$ be your favorite bounded sequence such that $\lim c_n$ does not exist. Let $a_n = c_n b_n$, then $a_n \to 0$ and $a_n/b_n = c_n$, and the limit does not exist.
But, don't consider this a disappointing result! We would be in a much different world if this was not the case; for example, imagine how boring L'Hopital's rule would be.
A: Let $a_n=1/n$, $b_n=1/n^2$. What can  you say on the ratio?
