Proving that the sequence $\left \{ \frac{x_{n}}{y_{n}} \right \} \rightarrow \frac{x}{y}$ Suppose $\left \{ x_{n} \right \}$ and $\left \{ y_{n} \right \}$ converge to the limits $x$ and $y$, respectively. Also, suppose that $y_{n}$'s are nonzero. I want to show that the sequence $\left \{ \frac{x_{n}}{y_{n}} \right \} \rightarrow \frac{x}{y}$.
Let $\epsilon >0$. We can pick an $N$ such that $n \geq N$ $\Rightarrow \left | \frac{x}{y} - \frac{x_{n}}{y_{n}}\right | < \epsilon$.
I start from $$ \left | \frac{x}{y} - \frac{x_{n}}{y_{n}}\right |$$ 
$$=\left | \frac{x}{y} - \frac{x}{y_{n}} + \frac{x}{y_{n}} - \frac{x_{n}}{y_{n}}\right | $$
$$= \left | \frac{x\left (y_{n}- y \right )}{y y_{n}} + \frac{\left ( x-x_{n} \right )}{y_{n}}\right | $$
$$\leq \left | \frac{x}{y} \right | \frac{1}{\left | y_{n} \right |} \left | y-y_{n} \right | + \frac{1}{|y_{n}|}\left | x-x_{n} \right | $$ by triangle inequality. 
I know that I want to make the two terms $\frac{\epsilon}{2}$ + $\frac{\epsilon}{2}$. Since $y_{n}$ converges to $y$, can make $\left | y - y_{n} \right |< \frac{\epsilon \left | y \right |}{\left ( \left | x \right | + 1\right )}$ for some $N_{1}$. 
But the $y_{n}$ term is giving me a problem, which I can't get rid off.
Any suggestions on how to proceed from here (or approaching the problem from a different angle) would be greatly be appreciated. 

You can proof
$$\lim_{n\to\infty}\frac{1}{y_n}=\frac{1}{y}$$
and then use the product rule.
