Let $x_n$ be an upper bounded sequence, $y_n$ is a sequence that $\forall n \in \mathbb{N}$ such that $y_n \rightarrow y>0 $ then prove that:
$$ \lim \sup \frac{x_n}{y_n} = \frac{\lim \sup x_n}{\lim y_n } $$
I don't know how to begin. Well I know that there exists an $M$ such that $x_n < M$ and that $y_n $ converges to $y$. This can be written with the $\epsilon - \delta$ definition but I don't know if this works.
Any suggestions, hints?
For sufficiently high $n$, $y_n>0$ so define $u_n:=\sup_{k>n}\frac{x_k}{y_k}$ for large $n$. This sequence is monotone so it has a limit. If $x=\limsup x_n$, we want to show that $\lim_{n\to\infty}u_n=\frac{x}y$. Fix $\epsilon>0$. Since $y_n\to y>0$ and $t\mapsto t^{-1}$ is continuous, $\frac1{y_n}\to\frac1y$. So there exists $N\in\mathbb{N}$ such that $$\frac1y-\epsilon<\frac1{y_k}<\frac1y+\epsilon$$ for all $k>N$. This implies $$\left(\frac1y-\epsilon\right)\sup_{k>n}x_k<u_n<\left(\frac1y+\epsilon\right)\sup_{k>n}x_k$$ for all $n\ge N$. Letting $n\to\infty$ we get $$\left(\frac1y-\epsilon\right)x\le\lim_{n\to\infty} u_n\le\left(\frac1y+\epsilon\right)x$$ by the squeeze law. The result follows by letting $\epsilon\to0$.
EDIT: Throughout I have implicitly assumed $x>-\infty$. If this is not the case then the only change we need to make is noting that $$\lim_{n\to\infty}u_n\le\left(\frac1y+\epsilon\right)\sup_{k>n}x_k=-\infty$$ and certainly $\frac{\limsup x_n}{\lim y_n}=-\infty$.