# Epsilon-Delta proof of non-existence of a limit

I know when you wanna prove that the limit of some function like $f(x)$ at $a$ exists you must prove a proposition of the following form:

$$(\forall \varepsilon >0)(\exists \ \delta >0)(0<|x-a|<\delta \ \Rightarrow \ |f(x)-l|<\varepsilon )$$

But what if you wanna prove that the limit of $f$ at $a$ does not exist. Do I need to negate the above expression and then reach it by recursive reasoning? Then what really is the negation of this expression? Should it lead to non-uniqueness of $l$?

Moreover I know that "Infinite limits" -which are limits with a value of infinity- are also classified as non-existent limits. Then how could we prove they don't exist using $\varepsilon$ and $\delta$?

Thank you in advance.

• You need to show that $\exists \epsilon>0, \forall \delta>0, \exists x_1, x_2 : |x_n-a|<\delta \text { and } |f(x_1)- f(x_2)| > 2\epsilon$ Jan 12 '17 at 19:24
• An example Jan 12 '17 at 19:36

For example, if $f$ is the function $x \mapsto {1\over x}$ as a function $f:(0,1) \to \mathbb R$ and you want to show that the limit towards $0$ does not exist then you might start by picking $\varepsilon = 1$.
Then your goal is to show that for every $\delta$ you can find $x\in (-\delta, \delta)$ such that $|f(x)|>1$.
So let $\delta>0$. Then you can find $n \in \mathbb N$ with ${1\over n}<\delta$ and $n>1$. Letting $x={1\over n}$ you get $|f({1\over n})| = n > 1$.