Spivak - Definition of the Converse of a Limit Spivak offers this as the definition of a limit: 
“The function “f” approaches the limit “l” near “a” means: for every e>0 there is some g>0 such that, if 0 < l x-a l < g, then l f(x)-I l < 0,”
and this as its converse: 
“there is some e > 0 such that for every g > 0 there is some x which satisfies 0 < l x-a l but not l f(x)-I l < e.” 
(Forgive my lack of “latexing” as I don’t have the program.) What are some examples of functions that the converse defines? How would you prove that the sum/product of two functions of this type (ie, whose limit doesn't exist) exists?
 A: Like Eevee says, you are talking about the negation of this definition.  But here is an example.  
Let
$$f(x) = \begin{cases} -1 & \text{if $x < 0$} \\ 1 & \text{if $x \geq 0$} \end{cases}$$
We claim $\lim_{x\to 0} f(x) =1$ is not true.
Let $\epsilon = \frac{1}{2}$.  Given $\delta > 0$, let $x = -\frac{\delta}{2}$.  Then $|x| < \delta$ by construction, but $f(x) = -1$, which is not in the interval $\left(\frac{1}{2},\frac{3}{2}\right)$.

A limit “not existing” is an even more complicated definition because it requires one more quantifier (four instead of three for the limit):

$\lim_{x\to a} f(x)$ does not exist if for all $L$, there exists $\epsilon > 0$ such that for all $\delta > 0$, there exists $x$ such that $0<|x-a| < \delta$ but $|f(x) - L| \geq \epsilon$.

We claim that the function $f$ above does not have a limit at $0$.  Let $L$ be given.  At least one of $L>-1$ or $L< 1$ must be true:


*

*If $L > -1$, there exists $\epsilon > 0$ such that $L-\epsilon > -1$.  Given $\delta > 0$, let $x = -\frac{\delta}{2}$.  Then $|x| < \delta$ but $f(x) < L -\epsilon$.  

*On the other hand, if $L < 1$, there exists $\epsilon$ such that $L+\epsilon < 1$.  Given $\delta > 0$, let $x = \frac{\delta}{2}$.  Then $|x|< \delta$ but $f(x) > L + \epsilon$.  

You also asked how one might prove that a sum of functions that don't have a limit could still have a limit.  The basic idea is that the combination might “cancel” the problematic part.  For instance, given $f$ as above, we could let $g = 3-f$. Then $g$ doesn't have a limit either, but their sum is the constant function $3$, and that obviously does have a limit.
