# The limit of a limsup of a nonnegative sequence is 0

I am currently looking at a result that states, for some nonnegative function $$g: \mathbb{R} \times \mathbb{R} \to [0,1]$$, there exists a constant $$c> 0$$ such that \begin{align} \lim_{k \to \infty} \limsup_{n \to \infty} |g(n,k) - c | = 0. ~(*) \end{align} It seems to me that the $$\limsup$$ is unnecessary, and could have been replaced by a $$\lim$$. This is because I believe that the above equation states that $$\forall \epsilon >0, \exists K_{\epsilon}$$ such that $$\forall k > K_{\epsilon}, \exists N_{k, \epsilon}$$ such that \begin{align} \sup_{n > N_{k,\epsilon}} |g(n,k) - c| < \epsilon. \end{align} (I found a nice explanation of the meaning of a double limit in quantifiers here Doubt about double limit definition. ).

Doesn't the $$\sup$$ imply that $$|g(n,k) - c| < \epsilon|$$ for all $$n > N_{k,\epsilon}$$, and vice versa? Also, is there any difference between $$(*)$$ and the following: \begin{align} \lim_{k \to\infty} \liminf_{n \to \infty} \frac{g(n,k)}{c} = \lim_{k \to\infty} \limsup_{n \to \infty} \frac{g(n,k)}{c} = 1 \end{align} ? Thanks so much.

• As an answer to my first question, it is noted in the comments of @Jack 's answer to math.stackexchange.com/questions/2420691/… that the \epsilon-\delta formulation of the repeated limit is not a definition but a consequence, and it does not imply the existence of the interior limit. Commented Nov 26, 2020 at 2:09

Consider $$g(n,k)=\frac12+\frac1{k^2+5} \cos(\pi n)$$ and $$c=\frac12$$. It satisfies the condition, but $$\lim_{n\to\infty}\lvert g(n,k)-c\rvert$$ does not exist for any $$k$$.