MHB 241.19 the e d definition of a limit.

[karush]

prove the statement using the $\epsilon,\delta$ definition of a limit.
$$\lim_{{x}\to{1}}\frac{2+4x}{3}=2$$
so then
$$x_0=1\quad f(x)=\frac{2+4x}{3}\quad L=2$$
now
$$0<|x-1|<\delta\quad\text
{and}\quad\left|\frac{2+4x}{3}-2\right|
<\epsilon$$
then
$$\left|\frac{2+4x}{3}-\frac{6}{3}\right|=\left|\frac{4x-4}{3}\right|$$
$$=\frac{4}{3}|x-1|=|x-1|<\frac{3}{4}\epsilon$$
finally
$$\left|\frac{2+4x}{3}-2\right|
=\frac{4}{3}|x-1|<\frac{4}{3}\delta
=\frac{4}{3}\left(\frac{3}{4}\epsilon\right)
=\epsilon.$$

 

[Opalg]

prove the statement using the $\epsilon,\delta$ definition of a limit.
$$\lim_{{x}\to{1}}\frac{2+4x}{3}=2$$
so then
$$x_0=1\quad f(x)=\frac{2+4x}{3}\quad L=2$$
now
$$0<|x-1|<\delta\quad\text
{and}\quad\left|\frac{2+4x}{3}-2\right|
<\epsilon$$

That last statement should read

Given $\epsilon>0$, we need to find $\delta>0$ such that $\left|\dfrac{2+4x}{3}-2\right| <\epsilon$ whenever $0<|x-1|<\delta$.​

then
$$\left|\frac{2+4x}{3}-\frac{6}{3}\right|=\left|\frac{4x-4}{3}\right|$$
$$=\frac{4}{3}|x-1|$$

That calculation should then continue $$\left|\frac{2+4x}{3}-\frac{6}{3}\right|=\left|\frac{4x-4}{3}\right|$$
$$=\frac{4}{3}|x-1| < \frac43\delta.$$ In order for that to be less than $\epsilon$, we can take $\delta = \dfrac34\epsilon.$

So your solution is essentially correct, but the wording needs to be clarified.