Answer
(a) $$\lim_{x\to0^+}\frac{2}{3x^{1/3}}=\infty$$
(b) $$\lim_{x\to0^-}\frac{2}{3x^{1/3}}=-\infty$$
Work Step by Step
(a) $$A=\lim_{x\to0^+}\frac{2}{3x^{1/3}}$$
As $x\to0^+$, $3x^{1/3}$ approaches $0$ from the right, where $3x^{1/3}\gt0$, so $2/3x^{1/3}$ will approach $\infty$. Therefore,
$$A=\infty$$
(b) $$B=\lim_{x\to0^-}\frac{2}{3x^{1/3}}$$
As $x\to0^-$, $3x^{1/3}$ approaches $0$ from the left, where $3x^{1/3}\lt0$, so $2/3x^{1/3}$ will approach $-\infty$. Therefore,
$$B=-\infty$$
