Answer
When $\epsilon = 0.2,~~~$ let $~~~\delta = 0.02$
When $\epsilon = 0.1,~~~$ let $~~~\delta = 0.01$
Work Step by Step
$f(x) = x^3-3x+4$
Let $\epsilon = 0.2$
$f(1.98) = (1.98)^3-3(1.98)+4 = 5.82$
$f(2.02) = (2.02)^3-3(2.02)+4 = 6.18$
Let $\delta = 0.02$
Then:
If $0 \lt \vert x-2 \vert \lt \delta,~~$ then $~~0 \lt \vert f(x)-6 \vert \lt \epsilon$
Let $\epsilon = 0.1$
$f(1.99) = (1.99)^3-3(1.99)+4 = 5.91$
$f(2.01) = (2.01)^3-3(2.01)+4 = 6.09$
Let $\delta = 0.01$
Then:
If $0 \lt \vert x-2 \vert \lt \delta,~~$ then $~~0 \lt \vert f(x)-6 \vert \lt \epsilon$
