#### Answer

please see step-by-step

#### Work Step by Step

By definition , $\displaystyle \lim_{x\rightarrow 2}(14-5x)=4$ is true if
for any $\epsilon > 0$, there is a $\delta > 0$ such that
if $0 < |x-2| < \delta $, then $|(14-5x)-4| < \epsilon$.
Analyzing the last expression,
$|(14-5x)-4| < \epsilon$
$|-5x+10| < \epsilon$
$|-5||x-2| < \epsilon$
$|x-2| < \displaystyle \frac{\epsilon}{5}$.
So, for any given $\epsilon$,
if we choose $\displaystyle \delta=\frac{\epsilon}{5}$,we will have
$ 0 < |x-2| < \delta \Rightarrow |(14-5x)-4|$