calculus - Negation of the Definition of Limit of a Function

Question

Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

and show this by the symbolism

$$\mathop {\lim f(x)}\limits_{x \to a} = L$$

Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from

$$\neg \left[ {\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)} \right]$$

and then go through it to get the final form of negation (See the example below).

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My Thought

I just wrote down the two following negations without going through a step by step approach.

$$\forall L,\exists \varepsilon > 0:\left( {\forall \delta > 0,\exists x:0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$

or

$$\nexists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

I want to know weather these are true or not.

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Example

I will give an example of what I mean by a step by step approach. Consider the following statement

$${P}\implies R$$

I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic

\begin{align}
\, \neg \left( {P \implies R} \right) &\iff \neg \left( {\neg P \vee R} \right) & \text{Conditional Disjunction} \\
\qquad \qquad \quad &\iff \neg \neg P \wedge \neg R & \text{Demorgan's Law} \\

\qquad \qquad \quad &\iff P \wedge \neg R & \text{Double Negation}
\end{align}

Answer

The first negation is almost completely right. You forgot to negate the implication at the end.

Remember that the negation of an "if, then" statement is not an "if, then" statement. $A \implies B$ has negation "$A \land \neg B$" (read: $A$ and not $B$).

So, "if the sky is blue, then I love cheese" has negation "the sky is blue and I do not love cheese."

We say $\lim \limits_{x \to a} f(x) = L$ if $$\forall \epsilon > 0\text{, }\exists \delta > 0 \text{ such that }\forall x\text{, }|x - a| < \delta \implies |f(x) - L| < \epsilon.$$ Then the negation of this is: $$\exists \epsilon > 0\text{ such that }\forall \delta > 0\text{, }\exists x\text{ such that }|x - a| < \delta \text{ **and** }|f(x) - L| \geq \epsilon.$$

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UPDATE Here is how to negate the following statement step by step

Negation of "$\lim \limits_{x \to a} f(x)$ exists", i.e., $$\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| \geq \epsilon).$$

We say $\lim \limits_{x \to a} f(x)$ does not exist if:

$\neg[\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L \neg[\forall\epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\neg[\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\forall \delta > 0\neg[\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: \neg[|x - a| < \delta \implies |f(x) - L| < \epsilon]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land \neg(|f(x) - L| < \epsilon)$ (Negation of implication)

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$