# If $\displaystyle\lim_{n\to \infty} x_n=a\neq 0$ and $\displaystyle \lim_{n\to \infty} x_ny_n=b$ then $\displaystyle \lim_{n\to \infty} y_n=b/a$?

I just came across the following analysis problem:

Let $a\neq 0$ and $(x_n)_{n\in\mathbb N}$ be a sequence of real numbers such that $\displaystyle\lim_{n\to \infty}x_n=a$. Let $(y_n)_{n\in\mathbb N}$ be another sequence of real number such that $\displaystyle\lim_{n\to \infty}x_ny_n=b$. Then I must show $\displaystyle\lim_{n\to \infty} y_n=\frac{b}{a}$.

I proceed as follows: \begin{align*} \displaystyle \left|y_n-\frac{b}{a}\right|&=\left|\frac{a}{a}y_n-\frac{b}{a}\right|\\ &=\frac{1}{|a|}|ay_n-b|\\ &=\frac{1}{|a|}|(ay_n-x_ny_n)+(x_ny_n-b)|\\ &\leq \frac{1}{|a|}(|ay_n-x_ny_n|+|x_ny_n-b|)\\ &=\frac{1}{|a|}(|y_n||x_n-a|+|x_ny_n-b|) \end{align*}

If I assume the sequence $(y_n)_{n\in\mathbb N}$ is bounded then the proof can be finished easily.

So, I'm in doubt if the result really holds without further assumptions on $(y_n)_{n\in\mathbb N}$.

Am I taking a wrong way or the boundedness is really needed?

The hypotheses guarantee that $\{y_n\}$ is bounded. Since the $\{x_n\}$ converge to a nonzero value, for suitably large $n$ you can force $|x_n| > C$ for some $C$ (such as $|a|/2$). Thus $|x_ny_n| \ge C|y_n|$ for large enough $n$. If the $y_n$ were unbounded this would imply that the $x_ny_n$ were unbounded, contradicting the assumption that $\{x_ny_n\}$ converges to $b$.
I would use the limit definition. $$\lim_{n\to\infty} x_n y_n \implies |x_ny_n-b| < \epsilon$$ for some $n>N$. We can rearrange that to show that $$|y_n| < \frac{\epsilon - b}{x_n}$$ which shows that the $$\lim_{n\to\infty}y_n$$ exists.