real analysis - $(y_n)$ bounded sequence and $limfrac{(x_n)}{(y_n)}=0$ show $lim(x_n) = 0$

If $(x_n)$ and $(y_n)$ are positive real sequences such that $(y_n)$ is bounded and $\lim\frac{(x_n)}{(y_n)} = 0$, prove that $\lim(x_n) = 0$

So since $(y_n)$ is bounded, there exists a real number $M$ such that for all n natural numbers, $(y_n) < |M|$
So by the limit definition :
$\lim|\frac{(x_n)}{(y_n)}-0| = \lim\frac{(x_n)}{(y_n)} < \epsilon$
So $|(x_n)| < \epsilon(y_n)$

Now I need to find a natural number $N$ such that for all $n > N$, $|(x_n)| < \epsilon$ is this correct which would prove this statement. However I'm having trouble picking something that will work here? Am I on the right track? I have to be very careful also about anything I use to quote the theorem since this is an introduction to real analysis class.

Thanks!

Answer

I would approach this by contradiction:

If the sequence does not converge to 0, then there is a subsequence that is bounded away from 0. So we may fix $a>0$ such that $|x_n|>a$ for infinitely many values of $n$.

Since the sequence of $y_n$ is bounded, say by $M$, we have $|y_n|a/M$.

This inequality holds for infinitely many $n$, and we have found a subsequence of $(x_n/y_n)$ that does not converge to 0.

Does this make sense?