Epsilon-delta proof of If $\lim (x_n y_n) = z$, then $\lim(y_n) = \frac{z}{x}$

Question

I would like to prove the statement

Let $(x_n)$ and $(y_n)$ be convergent sequence with limit $x$ and $y$ respectively. If $x_n >1$ for all $n$ and $\lim(x_n y_n) = z$, then $$\lim(y_n) = \frac{z}{x}$$

My attempt: I observed that $$\vert y_n - \frac{z}{x} \vert = \vert \frac{xy_n-z}{x} \vert = \frac{1}{|x|}\vert xy_n-z \vert$$

From the assumption, I believe that I should get $|x_n -1|$ somewhere. But I can't made it from the equation above...

Answer 1

A very short sketch of the proof may be given as follows,

Observe that (assuming $Y$ to a non-zero the bound of $(y_n)$), \begin{align}\left|x_ny_n-xy\right|&=\left|x_ny_n-xy_n+xy_n-xy\right|\\&=\left|y_n(x_n-x)+x(y_n-y)\right|\\&\le\left|y_n(x_n-x)\right|+\left|x(y_n-y)\right|\\&\le Y\left|x_n-x\right|+\left|x\right|\left|y_n-y\right|\\&< \color{crimson}{Y\cdot\dfrac{\varepsilon}{2Y}+|x|\cdot\dfrac{\varepsilon}{2|x|}}\\&<\varepsilon\end{align}for all sufficiently large $n$. From this it follows that $xy=z$ and consequently $y=\dfrac{z}{x}$.

Can you justify the coloured step?

Answer 2

$\lim\limits_{n\rightarrow\infty}|y_n-\frac{z}{x_n}|=\lim\limits_{n\rightarrow\infty}|\frac{x_ny_n-z}{x}|=\lim\limits_{n\rightarrow\infty}\frac{1}{|x_n|}|x_ny_n-z|=0$ since $x_ny_n\rightarrow z$ by assumption and $x_n>1\implies x_n\neq 0$.