Proving the lim as n goes to infinity of a function = 2

[Rubik]

I am unsure as to how to prove a that as the limit as n goes to inifinity of a certain function the answer is 2.. I am trying to use the definition of a limit and using $$\epsilon$$ and N - argument to get a contradiction in order to solve the equation.

 

[HallsofIvy]

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".