#### Answer

please see step-by-step

#### Work Step by Step

$\displaystyle \lim_{x\rightarrow a}f(x)=L $ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that the following is valid:
$($if $ 0 < |x-a| < \delta$ then $|f(x)-L| < \epsilon)$
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Given any $\epsilon > 0$, we want to find a $\delta > 0$ such that
$ 0 < | x-1| < \delta \ \ \Rightarrow\ \ |\displaystyle \frac{2+4x}{3}-2| < \epsilon$.
Analyzing $|\displaystyle \frac{2+4x}{3}-2| < \epsilon$,
we write equivalent inequalities:
$|\displaystyle \frac{2+4x}{3}-2| < \epsilon$
$|\displaystyle \frac{4x-4}{3}| < \epsilon$
$|\displaystyle \frac{4}{3}||x-1| < \epsilon \displaystyle \qquad/\times\frac{3}{4}$
$|x-1| < \displaystyle \frac{3}{4}\epsilon$.
The first and last inequalities are equivalent.
$\displaystyle \frac{3}{4}\epsilon$ is a positive nonnegative number, and we take it for our $\delta$.
So (equivalence being double implication),
$\displaystyle \delta=\frac{3}{4}\epsilon$,
$0