Answer
$$L=8$$ and $$\delta=\frac{1}{300}$$
Work Step by Step
1) From the definition of limit, the statement $$\lim_{x\to c}f(x)=L$$
means for each $\varepsilon\gt0$ there exists a $\delta\gt0$ such that if
$$0\lt|x-c|\lt\delta$$ then
$$|f(x)-L|\lt\varepsilon$$
2) $$\lim_{x\to2}(3x+2)$$
To find $L$, we plug $x=2$ into $f(x)$
$$\lim_{x\to2}(3x+2)=3\times2+2=8$$
So $L=8$.
Next, the question asks to find $\delta\gt0$ such that if $|f(x)-L|\lt0.01$ whenever $0\lt|x-c|\lt\delta$.
Here, $c=2$, $L=8$ and $f(x)=3x+2$.
Therefore, $$|x-c|=|x-2|$$
and $$|f(x)-L|=|3x+2-8|=|3x-6|=3|x-2|$$
Since $|f(x)-L|\lt0.01$, that means $3|x-2|\lt0.01$
Therefore as we need to find $\delta$ for $0\lt|x-2|\lt\delta$, we can pick $\delta=\frac{0.01}{3}=\frac{1}{300}$.
3) Try $\delta=\frac{1}{300}$ back again
For $0\lt|x-2|\lt\frac{1}{300}$, we have $$|f(x)-L|=3|x-2|\lt\Big(3\times\frac{1}{300}\Big)=\frac{1}{100}=0.01$$
So $\delta=\frac{1}{300}$ is a right choice.
