Negation of predicate logic/sequence of numbers The definition for limx→a f(x) = L is the following: 
For all real numbers epsilon > 0, there is a 
real number δ > 0 such that for all real numbers x if a − δ < x < a + δ and x not equal to a then 
L − epsilon < f(x) < L + epsilon. 
Write what it means for limx→a f(x) not equal to L. 
I'm trying to find the negation for what's stated above and I was wondering if this is right.
There is not a real number such that epsilon >0, for all real numbers 
 δ > 0 , There are real numbers x if a − δ < x < a + δ or x is equal to a then 
L − epsilon < f(x) < L + epsilon. 
I know I'm likely wrong...what's the right answer and how do you go about finding it?
 A: In the definition of limit, you should not have $\delta\lt x\lt a+\delta$. It is supposed to be $a-\delta\lt x\le a+\delta$.
Now we proceed to write out the definition of "It is not the case that $\lim_{x\to a} f(x)=L$."
The first thing to note is that in the definition of limit, we start by saying that for any $\epsilon \gt 0$, a certain thing happens. To negate it, we have to say that there exists an $\epsilon\gt 0$ for which the thing does not happen.
In the definition of limit, what thing happens for any given $\epsilon$? There is a $\delta\gt 0$ with certain properties. So in writing out the negation, we have to say there exists an $\epsilon \gt 0$ such that whatever $\delta\gt 0$ we pick, a certain thing will happen. 
What thing? There is an $x$ such that $a-\delta\lt x\lt a+\delta$ and $x\ne a$ and $f(x)$ is far from $L$. By far from $L$ we mean that $f(x)\le L-\epsilon$ or $f(x)\ge L+\epsilon$.
We could get at the same thing by formal manipulation of quantifiers, using the fact that $\lnot\forall t\phi$ is equivalent to $\exists t(\lnot \phi)$ and $\lnot \exists t \phi$ is equivalent to $\forall t(\lnot \phi)$.
