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
In terms of writing the logical negation, it is best to use a universal quantifier before the $x$: $$ \forall \varepsilon >0 \exists K>0: \forall x>K,\ \left | f(x)-l \right |<\varepsilon. $$ The logical negation of this statement is $$ \exists \varepsilon >0 \forall K>0: \exists x>K,\ \left | f(x)-l \right |>\varepsilon. $$ If you think about what this means, it means that you need to be able to find arbitrarily large $x$ with $|f(x)-l|$ bigger than a specified quantity.
For your concrete example, you can fix $\varepsilon=1/2$, say. Then, given any $K$, as the values of the sine range from $-1$ to $1$ in every interval of length $2\pi$, you can certainly find $x$ with $|\sin(x)-l|>1/2$, whatever number $l$ is.
Once you understand this, in practice what you need to do is find two sequences $\{x_k\}$, $\{y_k\}$, both going to infinity, such that $\{\sin x_k\}$ and $\{\sin y_k\}$ converge to different values.