Let:
$$
\begin{cases}
x_n = \frac{2-\cos\pi n}{2+\cos \pi n} \\
a = 3
\end{cases}
$$
Prove that $a$ is not the limit of $\{x_n\}$ using $\varepsilon$ definition.
Start with the definition of limit:
$$
\lim_{n\to \infty}x_n = a \stackrel{\text{def}}{\iff} \{\forall \varepsilon > 0, \exists N \in \mathbb N, \forall n>N:|x_n – a| < \varepsilon \}
$$
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 $k\in\mathbb N$:
$$
\begin{equation}
|x_n| =
\begin{cases}
3, & n=2k -1, \\
{1\over 3}, & n = 2k
\end{cases}
\end{equation}
$$
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:
- The negation of limit definition
- The proof itself
Btw is it true that any periodic sequence doesn't have a limit?
Thank you!
Best Answer
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.