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 $\lim _{n\to +\infty} x_n/y_n$?

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 $\epsilon = x/2 > 0$, there is an $n_2$ such that for all $n \geq n_2$ we have $$|x_n - x| < \epsilon$$ and therefore, $x_n > x/2 > 0$.

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 $n \geq n_0$: $$x_n/y_n > \epsilon \cdot M/\epsilon = M$$ and therefore $x_n/y_n \to + \infty$. Note that I assumed that $M > 0$. If $M \leq 0$, then taking $n_0 = \max\{n_1, n_2\}$, we have that $x_n/y_n > 0 \geq M$, so we really have that $x_n/y_n \to + \infty$.

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 $x_n/y_n$ does not have a limit: let $x_n = -1 + 1/n$ and $y_n = (-1)^n 1/n$, then $x_n/y_n = (-1)^{1-n}n + (-1)^{-n}$ 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.

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.

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 $\lim _{n\to +\infty} x_n/y_n$?

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 $\epsilon = x/2 > 0$, there is an $n_2$ such that for all $n \geq n_2$ we have $$|x_n - x| < \epsilon$$ and therefore, $x_n > x/2 > 0$.

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 $n \geq n_0$: $$x_n/y_n > \epsilon \cdot M/\epsilon = M$$ and therefore $x_n/y_n \to + \infty$. Note that I assumed that $M > 0$. If $M \leq 0$, then taking $n_0 = \max\{n_1, n_2\}$, we have that $x_n/y_n > 0 \geq M$, so we really have that $x_n/y_n \to + \infty$.

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 $x_n/y_n$ does not have a limit: let $x_n = -1 + 1/n$ and $y_n = (-1)^n 1/n$, then $x_n/y_n = (-1)^{1-n}n + (-1)^{-n}$ 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?

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 $\lim _{n\to +\infty} x_n/y_n$?

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 $\epsilon = x/2 > 0$, there is an $n_2$ such that for all $n \geq n_2$ we have $$|x_n - x| < \epsilon$$ and therefore, $x_n > x/2 > 0$.

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 $n \geq n_0$: $$x_n/y_n > \epsilon \cdot M/\epsilon = M$$ and therefore $x_n/y_n \to + \infty$. Note that I assumed that $M > 0$. If $M \leq 0$, then taking $n_0 = \max\{n_1, n_2\}$, we have that $x_n/y_n > 0 \geq M$, so we really have that $x_n/y_n \to + \infty$.

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 $x_n/y_n$ does not have a limit: let $x_n = -1 + 1/n$ and $y_n = (-1)^n 1/n$, then $x_n/y_n = (-1)^{1-n}n + (-1)^{-n}$ 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.