# Illustration of negation of limit at infinity

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.

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 $$2n\pi+\pi/6 > m$$