Illustration of negation of limit at infinity

Question

If we consider a function $f: R \to R$ and $L\in R$.

Then $L$ is said to be the limit of $f(x)$ as $x$ goes to infinity iff:

$\forall \epsilon \in R: [\epsilon \gt 0 \implies \exists m\in R: \forall x\in R: [m\lt x \implies |f(x)-L|\lt \epsilon]]$

And we can say that $L$ is NOT the limit of $f$ as $x$ goes to infinity iff:

$\exists \epsilon_0 \in R: [\epsilon_0 \gt 0 \land \forall m\in R: \exists x_0\in R: [m\lt x_0 \land |f(x_0)-L| \ge \epsilon_0]]$

I am trying to illustrate the negation with an example:

$f: R \to R$ such that $f(x)=\sin (x)$

If I consider $\epsilon_0 =1/2$, then there always exists $x_0 \in(2n\pi +\pi / 6, 2n\pi + 5\pi / 6) \cup (2n\pi +7\pi / 6, 2n\pi + 11\pi / 6) $ such that $\forall m (\gt 0) \in R : [m\lt x_0 \land |f(x_0)| \ge \epsilon_0]$ holds and thus the limit is not $L = 0$ as $x$ goes to infinity.

Am I going the right way? Please suggest.

Answer 1

All the ideas are there but you've made a small logical mistake, in that the quantity $n$ has not been specified.

After setting $\epsilon_0 > 1/2$, next you say:

Let $m > 0$.

Now you have to find a value of $x_0 > m$ such that $|f(x_0)| > 1/2$. To find that value of $x_0$ here's what you do:

Choose an integer $n$ such that $2n\pi+\pi/6 > m$. Now choose $x_0 \in (2 n \pi + \pi/6, 2 n \pi + 5 \pi / 6)$.

To be even more rigorous, you might want to replace that first sentence by this:

Applying the Archimedean principle, choose an integer $n$ such that $$n > \frac{m - \pi/6}{2\pi} $$ and therefore $$2n\pi+\pi/6 > m $$