Formal definition of limit does not exist what is the formal definition of:
The limit as $x$ goes to infinity of $f(x)$ does not exist
 A: $\lim_\limits {x\to a} f(x) = L$
If the limit exists, then $f(x)$ must meet the definition.
$\forall \epsilon > 0,\exists \delta > 0: |x-a|<\delta \implies |f(x) - L|<\epsilon$
If the limit does not exist then there is no $L$ such that satisfies the definition.
For any $L,$ there is an $\epsilon$ such that for any $\delta>0$ there is an $x$ with $|x-a|<\delta$ and $|f(x) - L| > \epsilon.$
When $x$ is going to infinity, $|x-\infty|<\delta$ isn't a construction that makes sense.  We tweak the definition as follows.
$\lim_\limits {x\to \infty} f(x) = L$
$\forall \epsilon > 0,\exists M > 0: x>M \implies |f(x) - L|<\epsilon$
If the limit does not exist.
For any $L,$ there is an $\epsilon$ such that for any $M>0$ there is an $x>M$ and $|f(x) - L| > \epsilon.$
Or, for any $M,$ there are $x,y>M$ where $f(x), f(y)$ are not arbitrarily close together.  Then we can choose $\epsilon =\frac {|f(y)-f(x)|}{2}$, and the definition cannot be satisfied.
A: It is instructive to write out the definition of limits at infinity in first-order logic. We then get
$\exists L. \forall \epsilon > 0. \exists m. \forall m'.
m' \geq m \Rightarrow |f(m')-L| \leq \epsilon$.
where $L$ denotes the limit at infinity that must exist. A negation of this inverts the quantifiers and the implication, and we get
$\forall L.\exists \epsilon > 0. \forall m. \exists m'. m' \geq m \wedge |f(m')-L| \geq \epsilon$
In other words, for every possible limit $L$ we can find a large $m'$ such that $f(m')$ is not close enough to $L$.
A: Intuitively, the limit of $f(x)$ is $L$ if "$f(x)$ gets as close as you like to $L$ given large enough $x$". Conversely, the limit of $f(x)$ is not $L$ if "$f(x)$ doesn't stay close enough to $L$ for large $x$". To say that the limit doesn't exist at all just means that no $L$ is the limit; so, the limit does not exist if "$f(x)$ doesn't stay close enough to anything for large $x$".
To say this precisely, the limit of $f(x)$ is $L$ if for every $\epsilon > 0$, there is a $y$ so that whenever $x > y$, $|f(x) - L| < \epsilon$ (that is, for any desired amount of closeness, there is a certain point after which $f(x)$ always stays that close to $L$).
The limit of $f(x)$ is not $L$ if this isn't true - that is, if there is some $\epsilon > 0$ so that for every $y$ there is an $x > y$ with $|f(x) - L| > \epsilon$. That is, for some desired amount of closeness, there never comes a point after which $f(x)$ stays that close to $L$ (after any point you like, $f(x)$ eventually gets too far away from $L$ again).
Finally, the limit of $f(x)$ does not exist if the above happens for every $L$ - so, if for every $L$ there is an $\epsilon > 0$ so that for every $y$ there is an $x > y$ with $|f(x) - L| > \epsilon$. That is, no matter which limit you try, there is some desired amount of closeness so that after any point you like, $f(x)$ eventually gets too far away from that limit again.
Wrapping it all in formal notation, this last one is: $(\forall L)(\exists\epsilon>0)(\forall y)(\exists x > y)|f(x) - L| > \epsilon$.
