Preliminaries: Suppose that \(\varepsilon>0,~(x^3-27)=(x-3)(x^2+3x+9)~ \&~|x-3|<1\) Then
\(-1<x-3<1\\2<x<4\\6<3x<12\\4<x^2<16\\10<x^2+3x<28\\19<x^2+3x+9<37\)
Now we build the limit. Suppose that \(\delta= \min\left\{1,\dfrac{\varepsilon}{37}\right\}\)
So if \(|x-3|<1\) then \(|x^2+3x+9|<37|\).
If \(|x-3|<\delta\) then it follows that \(\left|x^3-27\right|=\left|x-3\right|\cdot\left|x^2+3x+9\right|<\varepsilon\)