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