As long as I do not know the specific function, the only thing I can say is "use the definition".
To prove that \lim_{n\to\infty} f(n)= 2, show that, given any \epsilon> 0, there exist a number, N, such that if n>N, then |f(n)- 2|< \epsilon.
Again, exactly how you show that that is true depends upon the particular function. For example, if the function were
f(n)= \frac{2n}{n- 1}
Then I would need to make
\left|\frac{2n}{n-1}- 2\right|= \left|2-\frac{2}{n-1}- 2\right|= \left|\frac{2}{n-1}\right|< \epsilon
For n> 1, that is the same as
n- 1> \frac{2}{\epsilon}
or
n> 1+ \frac{2}{\epsilon}.
Now, for any given finite number, \epsilon, I can certainly choose a specific N and use that. Strictly speaking the proof would go the other way: Having chosen
N> 1+ \frac{2}{\epsilon}
if n> N, then
n> 1+ \frac{2}{\epsilon}
so that
n- 1> \frac{2}{\epsilon}
\frac{1}{n-1}< \frac{\epsilon}{2}
\frac{2}{n-1}< \epsilon
etc.
Because each step in working from
\left|\frac{2n}{n-1}- 2\right|< \epsilon
to
n> 1+ \frac{2}{\epsilon}
was invertible we can "work backwards" and so that is not typically shown. This is often referred to as "synthetic proof".
