Epsilon delta proof (exam type 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 $|x_n y_n -z|$ in the expansion above. But I can't made it

Hint:
First since $x_n>1,\forall n$, we hanve $x\ge 1$.
Following your proof, you have 
$$\frac{1}{|x|}| xy_n-z |=\frac{1}{|x|}| xy_n-x_ny_n+x_ny_n-z |\le \frac{1}{|x|}(| x-x_n||y_n|+|x_ny_n-z |)$$
Also note since $y_n$ is convergent, it's bounded. Then use convergence of $x_n$, $(x_ny_n)$
Note $y_n=\frac{x_ny_n}{x_n}$, in general you can show :
Given $\lim a_n=a,\lim b_n=b\not =0$, $b_n\not=0,\forall n$, we have $\lim \frac{a_n}{b_n}=\frac{a}{b}$. 
