[tex]\displaystyle\lim_{x\to2}(x^3-5x+5)=3[/tex]
means there exists some [tex]\delta[/tex] such that whenever [tex]0<|x-\delta|<2[/tex], it's guaranteed that [tex]|x^3-5x+5-3|<\epsilon[/tex] for all [tex]\epsilon>0[/tex].
You have
[tex]|x^3-5x+5-3|=|x^3-5x+2|=|(x-2)(x^2+2x-1)|=|x-2||x^2+2x-1|[/tex]
Completing the square for the quadratic term yields
[tex]|x-2||(x+1)^2-2|[/tex]
and by the triangle inequality,
[tex]|x-2||(x+1)^2-2|\le|x-2||x+1|^1+2|x-2|=|x-2|(|x+1|^2+2)[/tex]
Suppose we fix [tex]\delta\le1[/tex]. Then if [tex]|x-2|<\delta\le1[/tex], it follows that [tex]|x+1|\le4[/tex], since
[tex]|x-2|\le1\implies-1\le x-2\le1\implies 2\le x+1\le4[/tex]
This means
[tex]|x-2||(x+1)^2-2|\le|x-2|(|x+1|^2+2)\le|x-2|(4^2+2)=18|x-2|<\epsilon[/tex]
[tex]\implies|x-2|<\delta=\dfrac\epsilon{18}[/tex]
So to guarantee that [tex]x^3-5x+5[/tex] is within [tex]\epsilon[/tex] of the limit, we can choose the smaller of the two choices for [tex]\delta[/tex], or [tex]\delta=\min\left\{1,\dfrac\epsilon{18}\right\}[/tex].
Now, when [tex]\epsilon=0.2=\dfrac15[/tex], you would have [tex]\delta=\min\left\{1,\dfrac1{90}\right\}=\dfrac1{90}\approx0.0111[/tex], and when [tex]\epsilon=0.1=\dfrac1{10}[/tex], you would have [tex]\delta=\min\left\{1,\dfrac1{180}\right\}=\dfrac1{180}\approx0.0056[/tex].
