From the inequality defined by the limit$,$

$|4$x $12|<0\mathrm{.}5,$

we have the following.

$|$x $3|<0\mathrm{.}125$

We did so through the reversible operations of factoring and division. This means that the required inequality

$|4$x $12|<0\mathrm{.}5$

is true exactly when

$|$x $3|<0\mathrm{.}125.$

To finish, illustrate the precise definition of the limit

lim$$x $3(4$x $3)=9$

by finding the largest value of $$ corresponding to

$=0\mathrm{.}5.$

$$