Answer
$$L=4$$
and $$\delta=0.03$$
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) $$\lim_{x\to6}\Big(6-\frac{x}{3}\Big)$$
To find $L$, we plug $x=6$ into $f(x)$
$$\lim_{x\to6}\Big(6-\frac{x}{3}\Big)=6-\frac{6}{3}=6-2=4$$
So $L=4$.
Next, the question asks to find $\delta\gt0$ such that if $|f(x)-L|\lt0.01$ whenever $0\lt|x-c|\lt\delta$.
Here, $c=6$, $L=4$ and $f(x)=6-\frac{x}{3}$.
Therefore, $$|x-c|=|x-6|$$
and $$|f(x)-L|=\Big|6-\frac{x}{3}-4\Big|=\Big|2-\frac{x}{3}\Big|=\Big|\frac{6-x}{3}\Big|=\frac{|x-6|}{3}$$
(For absolute values, $|A|=|-A|$)
Since $|f(x)-L|\lt0.01$, that means $\frac{|x-6|}{3}\lt0.01$
Therefore as we need to find $\delta$ for $0\lt|x-6|\lt\delta$, we can pick $\delta=0.01\times3=0.03$.
3) Try $\delta=0.03$ back again
For $0\lt|x-6|\lt0.03$, we have $$|f(x)-L|=\frac{|x-6|}{3}\lt\frac{0.03}{3}=0.01$$
So $\delta=0.03$ is a right choice.
