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)
$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$
This proof is fine, but manipulating the bounded part can sometimes prove more challenging.
Here it is easy since it is only $|x||x-2|$, but often you will get benefits while shifting the limit towards $0$ by setting $x=x_0+u$ with $u\to 0$.
The main benefit is that limits in zero are part of your mathematical toolbox, also $|u|<\delta$ is easy to manipulate compared to $|x-x_0|<\delta$.
For instance here $|x(x-1)(x-2)-0|=\overbrace{|u|}^{<\epsilon}\overbrace{|1-u^2|}^{\le 1}<\epsilon$
Granted we take $\delta=\min(1,\epsilon)$ so that $\begin{cases}|u|<\delta<\epsilon\\0\le u^2<\delta^2<1^2\implies|1-u^2|\le 1\end{cases}$
The inequalities did appear more naturally than when you calculated the $10$ and $\frac 1{121}$ factors.
Another advantage you will encounter in your epsilon-delta proofs is how it is easy to bound polynomials.
For instance imagine that in the problem instead of $|1-u^2|$ you had $|1-u^2+7u^3-2u^4|$.
Just proceed using the fact that for $|u|<1$ then $\cdots<|u|^4<|u|^3<|u|^2<|u|<1$.
The triangle inequality gives you $$|1-u^2+7u^3-2u^4|<|1|+|u^2|+7|u^3|+2|u^4|<1+1+7+2=11$$
Conclude by taking $\delta=\min(1,\frac{\epsilon}{11})$.
Best Answer
Choose $\;\epsilon =1\;$ , and now we have that
$$|x+3-4|=|x-1|\ge 1\iff\begin{cases}x-1\ge 1\iff x\ge 2&,\;or\\{}\\x-1\le-1\iff x\le 0\end{cases}$$
Thus, for any $\;\delta >0\;$ , even if it is very small, take $\;x\ge 2\;$ s.t. $\;x-2<\delta\;$ , and then, as seen above, $\;|(x+3)-4|\ge 1\;\ldots$