Why does $|x-1|^2+3|x-1| < \epsilon \implies |x-1|^2 < \frac{\epsilon}{2} \ \ \land \ \ 3|x-1| < \frac{\epsilon}{2}$? I was reading some stuff about proving limits with the epsilon-delta definition from the exercises at http://www.vex.net/~trebla/homework/epsilon-delta.html. Anyway, as part of the solution to an exercise, we have this:
$$|x-1|^2+3|x-1| < \epsilon$$
$$ \implies |x-1|^2 \color{red}< \frac{\epsilon}{2} \ \ \land \ \  3|x-1| \color{red}< \frac{\epsilon}{2}$$

But, what is there to stop $|x-1|^2$ from being greater or equal than $\frac{\epsilon}{2}$?
The exact exercise is the one under "Quadratic Function".
 A: That isn't actually true. Choose $\epsilon = 100$, and let $x = 9$. Then:
$$
|9 - 1|^2 + 3|9 - 1| = 64 + 3 \cdot 8 = 88 < 100
$$
However, $|9 - 1|^2 = 64 > 100/2$.
So the statement is false.
The implication does work the other way, of course. Moreover, you can always fiddle with choosing your epsilons in some way to make continuity work (e.g. choose $\epsilon'$ such that $(\epsilon')^2 + 3\epsilon' < \epsilon$, etc.). But you have to be a bit more careful.
A: $|(x-1)^2|+|3(x-1)| \lt \epsilon$ does not imply that $|(x-1)^2| \lt \frac {\epsilon}2 \land |3(x-1)| \lt\frac {\epsilon}2 $. The text you linked to is actually telling you that it is the other way around. See the following.
We wish to prove that 
$$\forall \epsilon \gt 0, \; \exists \delta > 0 :$$
$$|x-1| \lt \delta \Rightarrow|(x-1)^2+3(x-1)| <\epsilon$$
The triangle inequality gives us
$$|(x-1)^2+3(x-1)| \le |(x-1)^2|+|3(x-1)|$$
We then have that 
$$|x-1| \lt \delta \Rightarrow |(x-1)^2| \lt \delta^2 \land |3(x-1)| \lt 3\delta$$
Let us now choose delta so 
$$max(\delta^2, 3\delta) \lt \frac {\epsilon}2$$
We then have 
$$|(x-1)^2+3(x-1)| \le |(x-1)^2|+|3(x-1)| \lt \delta^2 + 3\delta \lt \frac {\epsilon}2+ \frac {\epsilon}2 = \epsilon$$
A: The step you are referring to states:

$|(x-1)^2 + 3(x-1)| < ε$
  when  (several steps)
  $|(x-1)^2|<\frac{\varepsilon}{2}$ and $|3(x-1)|<\frac{\varepsilon}{2}$ 

which means 
$$ |(x-1)^2 + 3(x-1)| < ε \Leftarrow (x-1)^2<\frac{\varepsilon}{2} \text{ and } 3|x-1|<\frac{\varepsilon}{2}.$$
