Computing a Limit using the Limit Definition I've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.

Compute the limit. Then, using $\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. 
  $$\lim_{n \to \infty} \frac{1}{ln(ln(n))}$$

Firstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\infty}_{n=1}$ if for every $\epsilon > 0$ there is an integer $N=N(\epsilon)>0$ such that
$$|a_n - L|< \epsilon$$
for all $n\geq N$. Now for my informal interpretation. 
I believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\epsilon$? Any clarification of the definition would be appreciated.
Now, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that 
$$\left | \frac{1}{ln(ln(n))} - 0 \right | = \frac{1}{ln(ln(n))}$$
So now I need an $N$ such that, for all $n \geq N$, I get  $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that
$$\frac{1}{ln(ln(N))}<10^{-6}$$
So I just solve for N in the equality, which yields
$$N > e^{e^{1000000}}$$
Hence, for any $n \geq N$, 
$$\frac{1}{ln(ln(n))}<10^{-6}$$
Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated. 
 A: I think your confusion arises from the phrase "there exists an $N = N(\epsilon)$. You already showed why $N$ has to be a function of $\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.
You could just say then: choose $N > e^{e^{\epsilon^{-1}}}$ for $N(\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).
But, if you want to be explicit, you could use:


*

*$N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil$ where $\lceil - \rceil$ denotes the least integer greater than its argument, as you thought

*$N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil + 8434$

*$N(\epsilon) = \lceil 9434\pi e^{e^{\epsilon^{-1}}}\rceil$ 


See, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)
