Let $f(x)=\sqrt{\frac{\sin(x)}{x}}$.
Why isn't the $\lim\limits_{x\rightarrow \infty} f(x)$ equals to some $l \in \mathbb{R}$?
The definition of a finite limit at inifinity is:
$$\forall \epsilon>0, \exists X \in \mathbb{R}, \forall x \ge X,|f(x)-l| <\epsilon \tag{1}$$
It looks to me that $l=0$ meets the above definition...
I know that the negation of the definition of a finite limit at infinity is: $\exists \epsilon >0 , \forall X \in \mathbb{R}, \exists x \in \mathbb{R}, x\ge X \Rightarrow |f(x)-l| >=\epsilon \tag{2}$
Is $(2)$ true because, for any $X\in \mathbb{R}$, there will be an $x\ge X$ and $f(x)$ is undefined (i.e. the points where $\frac{\sin(x)}{x} < 0$) and we say the distance between $f(x)$ at these points and any $l\in\mathbb{R}$ is undefined and hence considered as greater than or equal to any $\epsilon > 0$? If yes, how do I write this out using the delta-epsilon proof? Otherwise, please tell me how to prove $(2)$
EDIT:
The definition for a finite limit and infinity according to my text is:
We say that $f(x)\rightarrow l$ as $x\rightarrow \infty$ for some real number,$l$, if:
for any $\epsilon>0$, there is an $X$ such that, for all $x\ge X$, $|f(x)-l|<\epsilon$
It could be that the definition of the limit requires that $f(x)$ is defined for all sufficiently large $x$. But to me, a more sensible definition could be written as $$\forall \epsilon>0, \exists X \in \mathbb{R}, \forall x \ge X\text{ with }x\in D_f,|f(x)-l| <\epsilon \tag{3}$$ where $D_f$ is the domain of $f$. With this definition, it must be an additional requirement that for any $X\in\mathbb{R}$, there are some $x\ge X$ with $x\in D_f$.
I think you will find that most mathematicians will go for this definition, whereas some elementary texts will go for the more restricted definition, in the interest of simplicity.