Doubt about double limit definition. Let $(a_n)_{n \in \mathbb{N}}$ a sequence. We say that the $\lim_{n \rightarrow \infty} a_n = L$, if for every $\varepsilon >0$ given, there exists $n_0 (\varepsilon) \in \mathbb{N}$, such that
$$n > n_{0}(\varepsilon) \Rightarrow |a_n - L| < \varepsilon $$
Consider $g: \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{R}$, I would like to know how I would write
$$\lim_{n \rightarrow \infty}\left( \lim_{k \rightarrow \infty} g(n,k)\right) = L $$
in terms of the $\varepsilon$-$\delta$ language.
My progress
First, I defined $\bar{g}(n)$ as $ \bar{g}(n) := \lim_{k\rightarrow \infty} g(n,k) $ . So fixing $n \in \mathbb{N}$, for every $\varepsilon/2 >0$ exists $k(n,\varepsilon) \in \mathbb{N}$, satisfying
$$k > k(n,\varepsilon) \Rightarrow |g(n,k) - \bar{g}(n)| < \varepsilon/2.$$
On the other hand $\lim_{n \rightarrow \infty} \bar{g}(n) = L$. Consequently for every $\varepsilon/2 >0,$ exists $n_0(\varepsilon) \in \mathbb{N},$ satisfying
$$n > n_0(\varepsilon) \Rightarrow |\bar{g}(n) - L| < \varepsilon/2 $$ 
Therefore, joining the two results above we have that the definition of the double limit would be:
For every $\varepsilon>0,$ exists $n_0(\varepsilon)$ $\in$ $\mathbb{N}$, and for every $n> n_0(\varepsilon)$, exists $k(n,\varepsilon)$ $\in$ $\mathbb{N}$, such that
$$n> n_0(\varepsilon),\hspace{0.1cm} k>k(n,\varepsilon) \Rightarrow |g(n,k) - L| < \varepsilon.  $$
Is this  correct? I'm not confident with my result.
Thanks in advance
 A: 
First, I defineddefine $\bar{g}(n)$ as 
  $$
 \bar{g}(n) := \lim_{k\rightarrow \infty} g(n,k). 
$$
   (I supposed that the limits always exists just to try to understand what is happening). So fixing $n \in \mathbb{N}$, for every $\varepsilon/2 >0$ exists $k_0(n,\varepsilon) \in \mathbb{N}$, satisfying
  $$
k > k_0(n,\varepsilon) \Rightarrow |g(n,k) - \bar{g}(n)| < \varepsilon/2.
$$
  On the other hand $\lim_{n \rightarrow \infty} \bar{g}(n) = L$. Consequently for every $\varepsilon/2 >0,$ exists $n_0(\varepsilon) \in \mathbb{N},$ satisfying
  $$
n > n_0(\varepsilon) \Rightarrow |\bar{g}(n) - L| < \varepsilon/2 
$$ 
Therefore, joining the two results above we have that the definition of the double limit would be:
For every $\varepsilon>0,$ there exists $n_0(\varepsilon)$ $\in$ $\mathbb{N}$, and for every $n> n_0(\varepsilon)$, there exists $k_0(n,\varepsilon)$ $\in$ $\mathbb{N}$, such that
  $$
n> n_0(\varepsilon),\hspace{0.1cm} k>k_0(n,\varepsilon) \Rightarrow |g(n,k) - L| < \varepsilon.  
$$

It looks good.

The limit in your question is called "repeated limit", not the "double limit". 
