[tex]\displaystyle\lim_{x\to1}(5x-3)=2[/tex]
is to say that for any [tex]\varepsilon>0[/tex], we can find [tex]\delta>0[/tex] such that [tex]0<|x-1|<\delta[/tex] implies [tex]|(5x-3)-2|<\varepsilon[/tex].
To that end, we have
[tex]|(5x-3)-2|=|5x-5|=5|x-1|<\varepsilon\implies|x-1|<\dfrac\varepsilon5=\delta[/tex]
which means that if [tex]\varepsilon=0.1[/tex] is given, then we can choose [tex]\delta=\dfrac{0.1}5=0.02[/tex]; if [tex]\varepsilon=0.05[/tex], then [tex]\delta=\dfrac{0.05}5=0.01[/tex]; if [tex]\varepsilon=0.01[/tex], then [tex]\delta=\dfrac{0.01}5=0.002[/tex].
