# $x_n \to x \in \mathbb{R}$ and $y_n \to 0$. What about $\frac{x_n}{y_n}$?

This is an exercise from my Analysis class.

Let $x_n, y_n$ be sequences in $\mathbb{R}$ with $x_n \to x \in \mathbb{R}\setminus \{0\}$ and $y_n \to 0$. Suppose that $y_n \neq 0$ for all $n \in \mathbb{N}$. What can we say about the existence and value of $\lim _{n\to +\infty} \frac{x_n}{y_n}$?

There is a hint: distinghuish two cases: $y_n$ has a fixed sign from some $n_0 \in \mathbb{N}$ onwards and the alternative.

In case $y_n$ has a fixed sign from some $n_0$ onwards, I think I have the following prove:

assume that $x_n \to x > 0$ and that $y_n$ is positive from some $n_1$ onwards. Let $M$ be an arbitrary real number. For $\epsilon = \frac{x}{2} > 0$, there is an $n_2$ such that for all $n \geq n_2$ we have $$|x_n - x| < \epsilon$$ and therefore, $x_n > \frac{x}{2} > 0$.

Because $y_n \to 0$, there is an $n_3 \in \mathbb{N}$ such that $|y_n| < M\epsilon$ for all $n \geq n_3$. Pick $n_0 = \max\{n_1, n_2, n_3\}$, then we have for $n \geq n_0$: $$\frac{x_n}{y_n} > \epsilon \cdot \frac{M}{\epsilon} = M$$ and therefore $\frac{x_n}{y_n}\to + \infty$. Note that I assumed that $M > 0$. If $M \leq 0$, then taking $n_0 = \max\{n_1, n_2\}$, we have that $\frac{x_n}{y_n} > 0 \geq M$, so we really have that $\frac{x_n}{y_n} \to + \infty$.

I think the other cases ($x_n$ has negative limit, $y_n$ is negative from some $n_0$ onwards, can be proven similary).

Question: if $y_n$ does not have a fixed sign from some $n_0$ onwards, I think the sequence  does not have a limit: let $x_n = -1 + \frac{1}{n}$ and $y_n = (-1)^n \frac{1}{n}$, then $\frac{x_n}{y_n} = (-1)^{1-n}n + (-1)^{-n}$ which has no limit. However, I have no real idea how to prove this. Any hints would be appreciated. Also, is my prove above correct?

Edit I suspect my course notes want a 'classic proof' using $n_0, \epsilon$ since this exercise appears after the part on classic results of converging sequences (sum, difference, quotient etc.). I have troubles with the fact that $y_n$ has no fixed sign from each $n_0$ onwards and how to use this in the proof.

• You can always take bijections $a:\mathbb{N}\to y^{-1}((0,\infty))$ and $b:\mathbb{N}\to y^{-1}((-\infty,0))$ and apply your argument above (which is correct) to $x_{a_n}/y_{a_n}$ and to $x_{b_n}/y_{b_n}$. On the other hand, the proof cannot be the analysis of the single example that you have at the end. You want to prove the general statement, that for every $y_n\to0$ without definite sign and $x_n\to x$, the limit of $x_n/y_n$ is not one of $\pm\infty$. – user574380 Jul 13 '18 at 20:00
• Thanks, but I am looking for a 'classic proof', using $n_0-\epsilon$. This question is one after basic results on convergent sequences (sum, difference, quotient rules etc.) – Student Jul 13 '18 at 20:03
• What I said is such a proof, because you will use your argument above, which is one. – user574380 Jul 13 '18 at 20:05

Assume $(y_n)_n$ doesn't eventually have a fixed sign. Therefore there are infinitely many positive terms and infinitely many negative terms. Let $(y_{p(n)})_n$ be the subsequence consisting of positive terms and let $(y_{q(n)})_n$ be the subsequence consisting of negative terms.
Then $(y_{p(n)})_n$ and $(y_{q(n)})_n$ have fixed signs so, assuming $x > 0$, we have $\lim_{n\to\infty} \frac{x_{p(n)}}{y_{p(n)}} = +\infty$ and $\lim_{n\to\infty} \frac{x_{q(n)}}{y_{q(n)}} = -\infty$ using the statement you already proved.
Therefore $\left(\frac{x_n}{y_n}\right)_n$ has two subsequences converging to different values so it cannot converge.