Answer
$L = -3, 𝛿 = 0.01$
Work Step by Step
Solve the inequality $|f(x) -c| < \epsilon$, where $c=-3$ and $\epsilon=0.02$:
$|3-2x+3| <0.02$
$|6-2x| <0.02$
$-0.02<6-2x<0.02$
$5.98<2x<6.02$
$2.990$ that places the open interval $(3-\delta, 3+\delta)$ centered at $3$ inside the interval $(2.99, 3.01)$. The distance between $3$ and $2.99$ is $0.01$ and distance between $3$ and $3.01$ is $0.01$. so we will take the smaller value for $\delta$, so
$\delta = 0.01$
The limit is $L=3$.
