<s>[INST] Prove that $\lim_{x\to 1}\left(4+x-3x^{3}\right)=2$. I was inspired by this <link> on page 10-11, where it was shown that $$\lim_{x\to 1}\left(4+x-3x^{3}\right)=2.$$
I wonder if it's possible to show this way without choosing a good value of $\delta>0$ (for example $\delta\leq 1$ in the link). Since 
$$\left | (4 + x-3x^{3})-2 \right |=\left | x-1 \right |\left | -3x^{2}-3x-2 \right |,$$
and if $\left | x-1 \right |<\delta $ then we have two inequalities
$$3(-\delta-1)<-3x<3(\delta-1)$$
and
$$-3(\delta+1)^{2}<-3x^{2}<-3(-\delta+1)^{2}.$$
Altogether we get
$$-3(\delta+1)^{2}+3(-\delta-1)-2<-3x^{2}-3x-2<-3(-\delta+1)^{2}+3(\delta-1)-2,$$
that is,
$$-9\delta-8-3\delta^{2}<-3x^{2}-3x-2<9\delta-8-3\delta^{2}.$$
Is it equivalent to 
$$\left | -3x^{2}-3x-2 \right |<9\delta+\left | -8-3\delta^{2} \right |=9\delta+8+3\delta^{2}?$$
If yes, then we finally have 
$$\left | x-1 \right |\left | -3x^{2}-3x-2 \right |<\delta(9\delta+8+3\delta^{2})$$
Now the question is: is there no way to isolate or solve the positive value of $\delta$ in the inequality, $$\delta(9\delta+8+3\delta^{2})<\epsilon?$$ If the solution is found, then I would prove this way:
Let $\epsilon>0$ be given. Choose $\delta>0$ such that $\delta <\textrm{something}$. If $0<\left | x-1 \right |<\delta$ then $$\left | (4 + x-3x^{3})-2 \right |=\left | x-1 \right |\left | -3x^{2}-3x-2 \right |<\dots<\epsilon$$
for all $x$ in the domain. 
 [/INST] Here is a solution along the lines you are suggesting:
Let $\epsilon>0$ be given, and choose $\delta>0$ with $\delta<\min\{\sqrt[3]{\frac{\epsilon}{9}}, \sqrt{\frac{\epsilon}{27}}, \frac{\epsilon}{24}\}$.
If $0<|x-1|<\delta$, then $\big|(4+x-3x^3)-2\big|=\big|2+x-3x^3\big|=\big|(x-1)(-3x^2-3x-2)\big|$
$=\big|x-1\big|\big|3x^2+3x+2\big|=\big|x-1\big|\big|3(x-1)^2+9(x-1)+8\big|\le\big|x-1\big|\big(3|x-1|^2+9|x-1|+8\big)$
$\;\;\;<\delta(3\delta^2+9\delta+8)=3\delta^3+9\delta^2+8\delta<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$.
</s>