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