Answer
1) $L=\lim_{x\to2}f(x)=2$
2) $\delta=1/3$
Work Step by Step
$$f(x)=\frac{4}{x}\hspace{1cm}c=2\hspace{1cm}\epsilon=0.4$$
1) Find $L=\lim_{x\to2}f(x)$ $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{4}{x}=\frac{4}{2}=2$$
Therefore, $L=\lim_{x\to2}f(x)=2$
2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-2|\lt\delta$ then $|f(x)-2|\lt0.4$
- First, solve the inequality to figure out the open interval on which the inequality holds.
$$|f(x)-2|\lt0.4$$ $$|\frac{4}{x}-2|\lt0.4$$ $$-0.4\lt\frac{4}{x}-2\lt0.4$$ $$1.6\lt\frac{4}{x}\lt2.4$$ $$\frac{4}{1.6}\gt x\gt \frac{4}{2.4}$$ $$2.5\gt x\gt\frac{5}{3}$$
The open interval on which the inequality holds is $(5/3,2.5)$.
- Find a value of $\delta\gt0$
$2$ is nearer to the endpoint $5/3$, whose distance is $1/3$.
So if we take $\delta=1/3$, then $0\lt|x-2|\lt1/3$, meaning all $x$ now are placed between $5/3$ and $2.5$, hence $|f(x)-2|\lt0.4$
In other words, $$0\lt|x-2|\lt\frac{1}{3}\Rightarrow|f(x)-2|\lt0.4$$
