More generally, let [tex]\delta=\frac\varepsilon4[/tex]. Then
[tex]|x-2|<\delta=\dfrac\varepsilon4\implies4|x-2|=|4x-8|=|(4x-2)-6)|<\varepsilon[/tex]
which proves the limit is 6. So
[tex]\varepsilon=0.5\implies\delta=0.125[/tex]
[tex]\varepsilon=0.1\implies\delta=0.025[/tex]
[tex]\varepsilon=0.05\implies\delta=0.0125[/tex]