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

1) The open interval is $(2-\frac{0.03}{m},2+\frac{0.03}{m})$ 2) $\delta=\frac{0.03}{m}$
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-2|\lt\delta\Rightarrow|mx-2m|\lt0.03\hspace{1cm}m\gt0$$ 1) Find the interval around $2$ on which $|mx-2m|\lt0.03$ holds. Solve the inequality: $$|mx-2m|\lt0.03$$ $$-0.03\lt mx-2m\lt0.03$$ $$-0.03\lt m(x-2)\lt0.03$$ $$-\frac{0.03}{m}\lt x-2\lt\frac{0.03}{m}\hspace{1cm}(m\gt0)$$ $$2-\frac{0.03}{m}\lt x\lt2+\frac{0.03}{m}$$ The open interval around $2$ is $(2-\frac{0.03}{m},2+\frac{0.03}{m})$. 2) Give a value for $\delta$ Both endpoints are equally distant from $2$, the distance of which is $\frac{0.03}{m}$ So if we take $\delta=\frac{0.03}{m}$ (since $m\gt0$, $\delta=\frac{0.03}{m}\gt0$) or any smaller positive number, then $0\lt|x-2|\lt\frac{0.03}{m}$, meaning all $x$ would be placed in the interval $(2-\frac{0.03}{m},2+\frac{0.03}{m})$ so that $|mx-2m|\lt0.03$. In other words, $$0\lt |x-2|\lt\frac{0.03}{m}\Rightarrow|mx-2m|\lt0.03$$