[tex]\displaystyle\lim_{x\to2}(x^3-3x+3)=5[/tex]
is to say that for any [tex]\varepsilon>0[/tex], we can find [tex]\delta>0[/tex] that guarantees
[tex]|x-2|<\delta\implies|(x^3-3x+3)-5|<\varepsilon[/tex]
[tex]|(x^3-3x+3)-5|=|x^3-3x-2|[/tex]
Observing that [tex]x^3-3x-2\bigg|_{x=2}=8-6-2=0[/tex], the polynomial remainder theorem tells us that we can factorize the cubic to get
[tex]|x^3-3x-2|=|(x-2)(x^2+2x+1)|=|(x-2)(x+1)^2|=|x-2|(x+1)^2[/tex]
If we assume [tex]0<\delta\le1[/tex], we can set up a corresponding upper bound on the quadratic factor. We start with [tex]|x-2|<\delta[/tex], from which we have
[tex]|x-2|<\delta\le1\implies1\le x-2\le3\implies4\le x+1\le6\implies(x+1)^2\le36[/tex]
Now,
[tex]|x-2|(x+1)^2<36|x-2|<\varepsilon\implies |x-2|=\dfrac\varepsilon{36}[/tex]
which suggests that we can choose [tex]\delta=\min\left\{1,\dfrac\varepsilon{36}\right\}[/tex] to ensure that we arrive at the inequality [tex]|(x^3-3x+3)-5|<\varepsilon[/tex].
So, given [tex]\varepsilon=0.2[/tex], we would have
[tex]\delta=\min\left\{1,\dfrac{0.2}{36}\right\}=\min\left\{1,\dfrac1{180}\right\}=\dfrac1{180}[/tex]
If [tex]\varepsilon=0.1[/tex], we would take
[tex]\delta=\min\left\{1,\dfrac{0.1}{36}\right\}=\min\left\{1,\dfrac1{360}\right\}=\dfrac1{360}[/tex]
