## University Calculus: Early Transcendentals (3rd Edition)

1) $L=\lim_{x\to-1}f(x)=1$ 2) $\delta=0.01$
$$f(x)=-3x-2\hspace{1cm}c=-1\hspace{1cm}\epsilon=0.03$$ 1) Find $L=\lim_{x\to-1}f(x)$ $$\lim_{x\to-1}f(x)=\lim_{x\to-1}(-3x-2)=-3\times(-1)-2=3-2=1$$ Therefore, $L=\lim_{x\to-1}f(x)=1$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-(-1)|\lt\delta$ then $|f(x)-1|\lt0.03$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-1|\lt0.03$$ $$|-3x-2-1|\lt0.03$$ $$|-3x-3|\lt0.03$$ $$-0.03\lt-3x-3\lt0.03$$ $$-0.03\lt-3(x+1)\lt0.03$$ $$0.01\gt x+1\gt-0.01$$ $$-0.99\gt x\gt-1.01$$ The open interval on which the inequality holds is $(-1.01,-0.99)$. - Find a value of $\delta\gt0$ The distance from $-1$ to both endpoints is the same, which is $0.01$. So if we take $\delta=0.01$, then $0\lt|x-(-1)|\lt0.01$, meaning all $x$ now are placed between $-1.01$ and $-0.99$, hence $|f(x)-1|\lt0.03$ In other words, $$0\lt|x-(-1)|\lt0.01\Rightarrow|f(x)-1|\lt0.03$$