Proof checking $\liminf cx_n = c \limsup x_n$

Question

I just started to work with $\limsup$'s and $\liminf$'s and I would like to know if my proof of the identity

\begin{equation} \liminf cx_n = c \limsup x_n \end{equation}

with $x_n$ a bounded sequence and $c\leq0$ is correct.

Let $a = \limsup x_n$ and $\epsilon>0$. Then

\begin{equation} x_n < a+\epsilon \end{equation}

for $n$ sufficiently large. Multiplying by $c$ we get the inequeality

\begin{equation} c x_n >ca + c\epsilon \end{equation}

or

\begin{equation} cx_n>ca-|c|\epsilon. \end{equation}

That is $\liminf cx_n = ca$ which implies $\liminf cx_n = c\limsup x_n$.

Answer 1

What you did only proves that $ca\leqslant\liminf_ncx_n$. You can prove that you actually have an equality by proving that some subsequence of the sequence $(cx_n)_{n\in\Bbb N}$ converges to $ca$ But that is easy. Take a subsequence $(x_{n_k})_{k\in\Bbb N}$ whose limit is $a$ and then $\lim_kcx_{n_k}=ca$.