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