Answer
1) $L=\lim_{x\to-3}f(x)=4$
2) $\delta=0.75$
Work Step by Step
$$f(x)=\sqrt{1-5x}\hspace{1cm}c=-3\hspace{1cm}\epsilon=0.5$$
1) Find $L=\lim_{x\to-3}f(x)$ $$\lim_{x\to-3}f(x)=\lim_{x\to-3}\sqrt{1-5x}=\sqrt{1-5\times(-3)}=\sqrt{1+15}=\sqrt{16}=4$$
Therefore, $L=\lim_{x\to-3}f(x)=4$
2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-(-3)|\lt\delta$ then $|f(x)-4|\lt0.5$
- First, solve the inequality to figure out the open interval on which the inequality holds.
$$|f(x)-4|\lt0.5$$ $$|\sqrt{1-5x}-4|\lt0.5$$ $$-0.5\lt\sqrt{1-5x}-4\lt0.5$$ $$3.5\lt\sqrt{1-5x}\lt4.5$$
Square: $$12.25\lt1-5x\lt20.25$$ $$11.25\lt-5x\lt19.25$$ $$-2.25\gt x\gt-3.85$$
The open interval on which the inequality holds is $(-3.85,-2.25)$.
- Find a value of $\delta\gt0$
$-3$ is nearer to the endpoint $-2.25$, whose distance is $0.75$.
So if we take $\delta=0.75$, then $0\lt|x-(-3)|\lt0.75$, meaning all $x$ now are placed between $-3.85$ and $-2.25$, hence $|f(x)-4|\lt0.5$
In other words, $$0\lt|x-(-3)|\lt0.75\Rightarrow|f(x)-4|\lt0.5$$
