Let $T\colon X\to X$ be continuous and $X$ compact and $K\subset X$ compact. By $s_n(2^{-k},K,T)$ denote the maximal cardinality of any $(n,2^{-k})$ separated subset of $K$.
Suppose, we know for given $n$ and $k$ that $s_n(2^{-k},K,T)$ is infinity and that this still holds when $k$ is the same but $n\to\infty$.
Then for some $c>1$ we shall have $$ s_n(2^{-k},K,T)\geq c^{kn}~~~(*) $$
The topological entropy $h(K,T)$ is given by $$ h(K,T)=\lim_{\varepsilon\to 0}\limsup_{n\to\infty}\frac{1}{n}\log s_n(\varepsilon,K,T). $$
My question is if it is then correct to use $(*)$ to say that $$ h(K,T)=\lim_{k\to\infty}\limsup_{n\to\infty}\frac{1}{n}\log s_n(2^{-k},K,T)\geq\lim_{k\to\infty}\limsup_{n\to\infty}\frac{1}{n}\log(c^{kn})=\lim_{k\to\infty} k\limsup_{n\to\infty}\log(c)=\infty $$
or if I am mixing something up with the limits.
I guess your main concern is whether $$ \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n, $$ given $a_n \leq b_n$. This is true since $$ \limsup_{n \to \infty} = \lim_{n \to \infty}\sup\{a_m : m \geq n\} \leq\lim_{n \to \infty} \sup\{b_m : m \geq n\} = \limsup_{n \to\infty} b_n. $$