First note that if [tex]x\neq1[/tex], you have
[tex]\dfrac{2x^4-6x^3+x^2+3}{x-1}=2x^3-4x^2-3x-3[/tex]
Now, you're looking for [tex]\delta>0[/tex] such that for any [tex]\varepsilon>0[/tex], you have
[tex]|x-1|<\delta\implies|2x^3-4x^2-3x-3+8|=|2x^3-4x^2-3x+5|<\varepsilon[/tex]
Note that you can divide through the left side of the [tex]\varepsilon[/tex] inequality by [tex]x-1[/tex] once more:
[tex]\dfrac{2x^3-4x^2-3x+5}{x-1}=2x^3-2x-5\implies 2x^3-4x^2-3x+5=(x-1)(2x^2-2x-5)[/tex]
So it follows that you need to find an appropriate [tex]\delta[/tex] that will guarantee
[tex]|(x-1)(2x^2-2x-5)|=|x-1||2x^2-2-5|<\varepsilon[/tex]
For the moment, let's fix [tex]\delta=1[/tex]. Then by this assumption, we have
[tex]|x-1|<\delta=1\implies-1<x-1<1\implies0<x<2[/tex]
From this we get
[tex]\implies0<x^2<4[/tex]
[tex]\implies0<x^2-x<4-2\implies 0<x^2-x<2[/tex]
[tex]\implies0<2x^2-2x<4[/tex]
[tex]\implies-5<2x^2-2x-5<-1[/tex]
[tex]\implies1<|2x^2-2x-5|<5[/tex]
where the upper bound is what we care about. With this assumption, we then get that
[tex]|x-1||2x^2-2-5|<5|x-1|<\varepsilon\implies|x-1|<\dfrac{\varepsilon}5[/tex]
which suggests that [tex]\delta[/tex] can be taken to be either the smaller of 1 or [tex]\dfrac{\varepsilon}5[/tex], or [tex]\delta=\min\left\{\dfrac{\varepsilon}5,1\right\}[/tex], to guarantee that the function gets arbitrarily close to -8.
