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: $$\exists \epsilon > 0\text{ such that }\forall \delta > 0\text{, }\exists x\text{ such that }|x - a| < \delta \text{ **and** }|f(x) - L| \geq \epsilon.$$
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)
$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$
Best Answer
Let's first see how we arrive at the correct negation using predicate logic and then justify: $$\neg \forall \epsilon > 0 \exists \delta > 0 \forall x, y \in I [ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon] \\ \exists \epsilon > 0 \neg \exists \delta > 0 \forall x,y \in I [ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon] \\ \exists \epsilon > 0 \forall \delta > 0 \neg \forall x,y \in I [ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon] \\ \exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I \neg [|x-y| < \delta \Rightarrow |f(x)-f(y)|<\epsilon] \\ \exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I [|x-y| < \delta \wedge \neg[|f(x)-f(y)|<\epsilon]] \\ \exists \epsilon > 0 \forall \delta > 0 \exists x,y \in I [|x-y| < \delta \wedge |f(x)-f(y)|\ge\epsilon]$$
What that means: There is some $\epsilon$ such that we can't bound the "change of $f$" around a point $x$ by $\epsilon$, no matter how close we stay to this $x$. Visualize this with $\tan x$. If $I = (-\frac\pi2,\frac\pi 2)$ points within $\delta$ of $-\frac\pi2$ or $\frac\pi2$ will be "bad" points where the implication $|x-y|<\epsilon \Rightarrow |\tan x - \tan y| < \delta$ fails.