Limit exists proof 
Let $x_n$ and $y_n$ be sequences in $\mathbb{R}$ such that
  $\lim(x_n)\ne 0$ and $\lim(x_ny_n)$ exists. Prove that the $\lim(y_n)$
  exists.

 A: Denote $a_n=x_n y_n.$ Then $y_n=\dfrac{a_n}{x_n}.$ 
If exists $\lim\limits_{n\to\infty}{x_n}\ne{0}$ then also exists $\lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\dfrac{1}{\lim\limits_{n\to\infty}{x_n}}.$  If, in addition, exists $\lim\limits_{n\to\infty}{a_n}=\lim\limits_{n\to\infty}{x_n y_n},$ then exists
$$\lim\limits_{n\to\infty}{y_n}=\lim\limits_{n\to\infty}{\left(a_n\cdot \dfrac{1}{x_n}\right)}=\lim\limits_{n\to\infty}{a_n}\cdot \lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\\
=\lim\limits_{n\to\infty}{(x_n y_n)}\cdot \lim\limits_{n\to\infty}{\dfrac{1}{x_n}}=\lim\limits_{n\to\infty}{(x_n y_n)}\cdot\dfrac{1}{\lim\limits_{n\to\infty}{x_n}}.$$ 
A: I give a proof for the case where $y_n$ is bounded. 
Suppose $y_n$ is bounded, then we can show that 
$\forall \epsilon > 0, \exists N_1$ s.t. $\forall n \geq N_1$, $|x_n y_n - x y_n| < \frac{\epsilon}{2} $
Denote $\lim x_n y_n = z$ and $\lim x_n = x$. 
Therefore $\forall \epsilon > 0, \exists N_2$ s.t. $\forall n \geq N_2$, $|x_n y_n - z| < \frac{\epsilon}{2} $
We have $|xy_n - z| \leq |xy_n -x_ny_n| + |x_ny_n - z|$.
Therefore for a given $\epsilon > 0$ we have $N = \max (N_1,N_2)$ such that $\forall n \geq N$
$|x y_n - z| < \epsilon$.
Since $x \ne 0$, this proves that $\lim y_n = \frac{z}{x}$.
A: Hint: You're told that $\lim (x_n)\ne 0$. What can you do to something which isn't $0$? Use the fact that whenever $\lim (a_n), \lim (b_n)$ exist, then $\lim (a_nb_n)$ exists.
Edit: Using the property that if $\lim (a_n), \lim (b_n)$ exist, then $\lim (a_nb_n)$ exists, let $\displaystyle a_n=\frac{1}{x_n}$ if $x_n\neq 0$ and $a_n=1$ is $x_n=0$, (note that $\lim (x_n)\neq 0$ does not imply that $x_n\neq 0$ for all $n\in \Bbb N$). Set $b_n=x_ny_n$. The limits $\lim (a_n), \lim (b_n)$ exist. You get 
$$\lim (a_n)\lim(b_n)=\lim(a_nb_n)=\lim\left(\frac{1}{x_n}\cdot x_ny_n\right)=\lim (y_n)$$
this tells you that $\lim (y_n)=\lim (a_n)\lim(b_n)\in \Bbb R$, which means $\lim (y_n)$ exists.
