Answer
$\lim\limits_{x\to0^+}(6-\dfrac{1}{x^3})=-\infty.$
Work Step by Step
$\lim\limits_{x\to0^+}(6-\dfrac{1}{x^3})\to$
$\lim\limits_{x\to0^+}(6)=6.$
$\lim\limits_{x\to0^+}\dfrac{-1}{x^3}=\dfrac{-1}{(0^+)^3}=\dfrac{-1}{0^+}=-\infty.$
By Theorem $1.15:$
$\lim\limits_{x\to0^+}(6-\dfrac{1}{x^3})=-\infty.$
