Htykuut

Let:
$$
begin{cases}
x_n = frac{2-cospi n}{2+cos pi n} \
a = 3
end{cases}
$$
Prove that $a$ is not the limit of ${x_n}$ using $varepsilon$ definition.

Start with the definition of limit:
$$
lim_{nto infty}x_n = a stackrel{text{def}}{iff} {forall varepsilon > 0, exists N in mathbb N, forall n>N:|x_n - a| < varepsilon }
$$

Now negate that definition. For the sequence to not have a limit we have the following to be true:

$$
{exists varepsilon >0, forall N in mathbb N, exists n > N: |x_n - a| ge varepsilon}
$$

Try to find such epsilon. Since $n in mathbb N$ we have that for $kinmathbb N$:

$$
begin{equation}
|x_n| =
begin{cases}
3, & n=2k -1, \
{1over 3}, & n = 2k
end{cases}
end{equation}
$$

Now take for instance $varepsilon = 1$. With that $varepsilon$ for any $N$ we may find infinitely many $n$ such that $n > N text{and} |x_n -3| ge 1$. Thus the sequence does not have a limit.

I'm kindly asking to verify two things:

1. The negation of limit definition


2. The proof itself

Btw is it true that any periodic sequence doesn't have a limit?

Thank you!

The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.