## Calculus 10th Edition

$$L=4$$ and $$\delta=0.03$$
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.