Answer
$$\lim_{x\to0^+}\frac{2}{x^{1/5}}=\infty$$
$$\lim_{x\to0^-}\frac{2}{x^{1/5}}=-\infty$$
Work Step by Step
(a) $$A=\lim_{x\to0^+}\frac{2}{x^{1/5}}$$
As $x\to0^+$, $x^{1/5}$ approaches $0$ from the right, where $x^{1/5}\gt0$, so $2/x^{1/5}$ will approach $\infty$. Therefore,
$$A=\infty$$
(b) $$B=\lim_{x\to0^-}\frac{2}{x^{1/5}}$$
As $x\to0^-$, $x^{1/5}$ approaches $0$ from the left, where $x^{1/5}\lt0$, so $2/x^{1/5}$ will approach $-\infty$. Therefore,
$$B=-\infty$$
