Prove that $(x_n)\rightarrow 0$ if $(y_n)\rightarrow 0$ I want to prove the following theorem:

Let $q$ be a real number such that $0 ≤ q < 1$. Let $(x_n)$ be a bounded sequence and let $(y_n)$ be the sequence defined by $∀n ∈ \mathbb{N}, y_n = x_n − qx_{n+1}$. If $y_n$ converges to $0$, then $x_n$ converges to $0$.

I can show that when $\lim x_n$ exists, $(x_n)\rightarrow 0$, but my trouble is showing that $\lim x_n$ in fact exists. My strategy is to show that $\limsup x_n = \liminf x_n$ so that I may conclude that $\lim x_n$ exists. Of course, when $q=0$, the result is immediate.
Here's what I have so far, when $0<q<1$,
\begin{align*}
\limsup y_n & = \limsup(x_n-qx_{n+1})\\
    & \leq \limsup(x_n) + \limsup(-(qx_{n+1})\\
& = \limsup(x_n) -\liminf(qx_{n+1})\\
& = \limsup(x_n) - q\liminf(x_{n+1})\\
& = \limsup(x_n) - q\liminf(x_n),
\end{align*}
and by similar reasoning I've been able to show 
\begin{align*}
\liminf y_n \geq \liminf x_n -q\limsup x_n.
\end{align*}
You can assume that each line follows from a result I proved in a previous problem. Given what I have so far, is there a way to reach $\limsup x_n \leq \liminf x_n$ so that I may conclude $\limsup x_n = \liminf x_n$? What am I missing?
I want to use the fact that $y_n$ converges to $0$, but whenever I try using this fact with my inequalities, I just arrive at what we already know: $\liminf x_n \leq \limsup x_n$.
 A: $0\le q<1$ then $-1<-q$ iff $-|x_{n+1}|\leq-q|x_{n+1}|$ iff $|x_n|-|x_{n+1}|\leq|x_n|-q|x_{n+1}|\leq|x_n-qx_{n+1}|=|y_n|\to0$
Therefore $|x_n|$ is a Cauchy's sequence. Suppose $|x_n|\to c$. Applying limit in $|x_n|-q|x_{n+1}|\leq|y_n|$ we obtain $(1-q)c=0$. As $1-q \neq0$, we get $c=0$, i.e., $|x_n|\to0$. Therefore $x_n\to0$.
A: Jaca's answer is fine but here's an argument following your line of thought in case you need it.
So, the fact that $(x_n)$ is a bounded sequences allows us to play with both $\limsup x_n$ and $\liminf x_n$ as numbers, what I want to say is that both limits are finite.
Remember that, since $\lim y_n=0$, we have both limits superior and inferior equal to zero.
We have $x_n=y_n+qx_{n+1}$, taking $\limsup$ on both sides we get
\begin{align}
  \limsup x_n &\leq \limsup y_n + q\limsup x_{n+1} \\
              &= \limsup y_n + q\limsup x_{n}
  .
\end{align}
So that
$$
  (1-q)\limsup x_n \leq 0
$$
Since $1-q>0$ (we're in the interesting case $0<q<1$) we must have $\limsup x_n\leq 0$. Now take $\liminf$ in the equation $x_n=y_n+qx_{n+1}$ so that
$$
  \liminf x_n \geq q\liminf x_n \quad\Rightarrow\quad (1-q)\liminf x_n\geq 0
  ,
$$
and similiarly, $1-q>0$ implies $\liminf x_n\geq 0$ and we obtain $\liminf x_n=\limsup x_n =0$.
