Answer
When $\epsilon = 0.1$, we can let $\delta = 0.02$
When $\epsilon = 0.05$, we can let $\delta = 0.01$
When $\epsilon = 0.01$, we can let $\delta = 0.002$
Work Step by Step
It is given that: $\lim\limits_{x \to 2}(5x-7) = 3$
We can let $\delta = \frac{\epsilon}{5}$
Let $\epsilon = 0.1$
Let $\delta = 0.02$
Suppose that $\vert x-2 \vert \lt \delta$
Then:
$\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.02) = 0.1$
Let $\epsilon = 0.05$
Let $\delta = 0.01$
Suppose that $\vert x-2 \vert \lt \delta$
Then:
$\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.01) = 0.05$
Let $\epsilon = 0.01$
Let $\delta = 0.002$
Suppose that $\vert x-2 \vert \lt \delta$
Then:
$\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.002) = 0.01$
