How to complete this epsilon delta proof Prove $\lim_{x\to 1} {2+4x \over 3} = 2$ using the epsilon delta definition of a limit.
if $0 < \left|x-1\right| < \delta$ then $\left|{2+4x \over 3}-2\right| < \epsilon$
scratch work for finding $\delta$
$$
\left|{2+4x \over 3}-2\right| < \epsilon
$$
$$
2-\epsilon < {2+4x \over 3} < 2 + \epsilon
$$
$$
6-3\epsilon < 2+4x < 6 + 3\epsilon
$$
$$
4-3\epsilon < 4x < 4 + 3\epsilon
$$
$$
1-\frac34\epsilon < x < 1 + \frac34\epsilon
$$
$$
\frac34\epsilon < x-1 < \frac34\epsilon
$$
$$
\left|x-1\right| < \frac34\epsilon
$$
This is where I need help.
Given $\epsilon > 0$ and let $\delta = \frac34\epsilon$
$$\left|{2+4x\over 3}-2\right| < \frac43\delta$$
$$-\frac43\delta < {2+4x \over 3}-2 < \frac43\delta $$
$$2-\frac43\delta < {2+4x\over 3}< 2 + \frac43\delta$$
$$6-4\delta<2+4x < 6+4\delta $$
$$4-4\delta<4x<4+4\delta$$
$$1-\delta<x<1+\delta$$
$$\left|x-1\right|<\delta$$
I'm pretty sure the $\frac43\delta$ part is wrong on the first line which makes the rest wrong but I'm not sure how to do this. Thanks.
 A: Your first part of your solution is correct.  You have indeed found a $\delta=\frac34\epsilon$ that will work.  Then you proceed like this.
$$\forall\epsilon>0,\ \exists\delta>0\ \text{such that }|x-1|<\delta\implies\left|\frac{2+4x}{3}-2\right|<\epsilon$$
Let $\delta=\frac34\epsilon$.  Then,
$$|x-1|<\delta\implies|x-1|<\frac34\epsilon\implies\frac43|x-1|<\epsilon\implies\left|\frac{4x-4}{3}\right|<\epsilon\quad\implies\left|\frac{2+4x-6}{3}\right|<\epsilon\implies\left|\frac{2+4x}{3}-2\right|<\epsilon\implies|f(x)-L|<\epsilon$$
And we are done.
A: Nothing is wrong with what you have done. For every $\epsilon>0$, you needed to find a $\delta > 0$ such that $0<|x-1|<\delta \implies |\frac{2+4x}{3}-2|<\epsilon$. Your first derivations (before you say "This is where I need help") actually find such a $\delta$ (as they are all "if and only if" statements). That is, you have shown $\delta = \frac{3}{4}\epsilon$ works. I suggest you take some time for your proof to "sink in" so as to clear any confusion that you may have.
