Doubt about double limit definition.

Question

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

Answer 1

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.

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