Answer
$\lim\limits_{x \to 1} \frac{2+4x}{3} = 2$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = \frac{3\epsilon}{4}$
Suppose that $\vert x-1 \vert \lt \delta$.
Then: $\vert \frac{2+4x}{3}-2 \vert = \vert \frac{2+4x-6}{3} \vert = \vert \frac{4x-4}{3} \vert = \frac{4}{3}\vert x-1 \vert \lt \frac{4}{3}\delta = \frac{4}{3}(\frac{3\epsilon}{4}) = \epsilon$
Therefore, $\lim\limits_{x \to 1} \frac{2+4x}{3} = 2$
